Class 12 Engineering Chemistry - The D-And F-Block Elements

The sum of the total number of bonds between chromium and oxygen atoms in chromate and dichromate ions is ________.

Total $$Cr-O$$ bonds in chromate $$4\sigma$$ and $$2 \pi$$ while in dichromate ion $$8\sigma$$ and $$4\pi$$ bonds.

Hence total $$Cr-O$$ bonds will be $$6+12=18$$

1476923_1697537_ans_99dc527776af4f02b0c48c7e57557510.png

An acidified solution of potassium chromate was layered with an equal volume of amyl alcohol. When it was shaken after the addition of $$1 \ mL$$ of $$3\% \ H_2O_2$$. a blue alcohol layer was obtained. The blue color is due to the formation of a chromium (VI) compound $$‘X’$$. What is the number of oxygen atoms bonded to chromium through only single bonds in a molecule of $$X$$?

$$CrO_4^{2-} + H_2O_2 + H^+ \rightarrow CrO_5 + H_3O^+$$

Number of oxygen atoms bonded to chromium through only single bonds $$= 4$$
1987539_1950108_ans_205f5b63fae943e78945200c7083a783.png

Explain why:
Transition elements form coloured compounds.

Transition metal ions generally possess one or more unpaired electrons. When visible light falls on a transition metal compound or ion, the unpaired electrons present in the lower energe d-orbital get promoted to high energy d-orbitals, called d-d transition, due to the absorption of visible light. Since, the energy involved in d-d transition is quantised, only a definite wavelength gets absorbed, remaining wavelengths present in the visible region got transmitted. Therefore, transmitted light shows some colour complementary to the absorbed colour.

................... gives blue colour to the Glass.

To impart colour to the glass, a trace amount of transition metal oxide is added.
CoO or CuO gives blue colour to the Glass.

Why do transition elements form complexes?

The cations of d-block elements have strong tendency to form complexes with certain molecules $$\left ( e.g. CO , NO,NH_{3} ... etc\right )$$ or ions $$\left ( e.g. F^{-}, Cl^{-}, CN^{-} \right )$$ called ligands. Their tendency to form complexes is due to two reasons.

Small size and high positive charge density of ions of transition metals.
Presence of vacant $$(n-1)d$$ orbitals which are of appropriate energy to accept one pair and unshared pair of electrons from the ligands for bonding with them. e.g. of some complex compounds are:

$$\left [ Cu\left ( NH_{3} \right )_{4} \right ]^{2+},\left [ Ag\left ( NH_{3} \right )_{2} \right ]^{^{+}},\left [ Fe\left ( CN \right )_{6} \right ]^{4-} ...$$etc.

Differentiate Lanthanides and Actinides.

Lanthanides	Actinides
1. Binding energies of 4f electrons are higher.	1. Bending energies of 5f electrons are lower.
2. Maximum oxidation state exhibited by lanthanides is +4 e.g. Ce$$^{4+}$$	2. Due to lower binding energies they show higher oxidation state such as +4, +5 and +6. Uranium exhibits +6 oxidation state in UF$$_6$$ and UO$$_2$$Cl$$_2$$.
3. 4f electrons have greater shielding effect.	3. 5f electrons have poor shielding effect.
4. Most of their ions are colourless	4. Most of their ions are coloured U$$^{3+}$$ (red), U$$^{4+}$$ (green) and UO$$_2^{2+}$$ (yellow)
5. They are paramagnetic but magnetic properties can be easily explained	5. They are also paramagnetic but their magnetic properties are very difficult to interpret.
6. They do not form complexes easily	6. They have much greater tendency to form complexes.
7. Except promethium, they are non-radioactive.	7. All of them are radio active.
8. Their compounds are less basic.	8. Their compounds are more basic.
9. They do not form oxocations	9. They form oxocations such as $$UO_2^{2+}, UO^+, NpO_2^+, PuO_2^+$$

For the dichromate dianion, which of the following is true?
1) $$4 Cr - O$$ bonds are equivalent
2) $$6 Cr - O$$ bonds are equivalent

The correct answer is - There are '6' identical $$Cr - O$$ bonds in the dichromate ion$$ (Cr_2O_2^{2-}).$$

In a dichromate ion $$ (Cr_2O_7^{2-}).$$, there are 6 terminal $$Cr – O$$ bonds. All these bonds are equivalent due to resonance.

The Dichromate ions $$ (Cr_2O_7^{2-}).$$ are oxoanions of chromium in the oxidation state +6.

Transition metals are good catalytic agent. Why?

Transition metals are good metal catalyst because they easily lend and take electrons from other molecules.

Represent the structure of dichromate ion.

The structure of dichromate ion is as shown. It has two tetrahedral units, each shares one corner. $$\displaystyle Cr-O-Cr$$ bond angle is $$\displaystyle 126^o$$

What is the magnetic property of $$Ni$$ atom $$[Z=28]$$ ?

1376523_1166480_ans_0f2e2755f228466a9367af1eb3230401.jpeg

Name the method to determine magnetic properties.

1376525_1166562_ans_f0dfcd2c2e53417683105650b12e1cb7.jpeg

How do you account for the larger size of lanthanum than hafnium though they are successive elements?

Lanthanum is the first element of $$3^{rd}$$ series of transition elements. The differentiating electron goes into '5d' subshell. Though hafnium is present next to lanthanum, its atomic number is 72. That means, there is a large difference in nuclear charge. Though 14 electrons are added to the inner shell, the poor screening effect of $$f$$ subshell is not sufficient to balance the increased nuclear charge. Increase in effective nuclear charge results in a decrease in atomic radii. This makes the atomic size of hafnium smaller than lanthanum.

Magnetic moment of $$V^{2+}$$ = Magnetic moment of $$Co^{2+} $$ if true enter 1, if false enter 0.

The correct answer is 1.

Magnetic moment of $$V^{2+} =$$ Magnetic moment of $$Co^{2+} =$$ 3.87 B.M

Both elements have 3 unpaired electrons.

$$\mu =\sqrt{(n(n+2)}$$

$$\mu = \sqrt{(3(3+2)}$$

$$\mu = 3.87$$ B.M.
Note: The electronic configuration of $$V^{2+}$$ is $$[Ar]3d^3$$
The electronic configuration of $$Co^{2+}$$ is $$[Ar]3d^7$$

The value of 3.00 BM into SI unit is $$x \times 3.09 \times 10 ^{-24} Am^2$$. The value of $$x $$ is ______.

1 BM is equal to $$9.27 \times 10^{-24}\ Am^2$$

Therefore, 3 BM = $$3 \times 9.27 \times 10^{-24}$$ = $$ x \times 3.09 \times 10^{-24}$$

$$x = 9$$.

The colour of $$MnO^-_4$$ ion is due to charge transfer.
If true enter 1, else enter 0.

$$M n O^-_4$$
Here, $$Mn$$ is in $$+7$$ oxidation state.

The electronic configuration of Mn is $$[Ar]4 s^2 3d^5$$. Mn in its +7 oxidation state means $$d^0$$ electron. Its colour is therefore, due to charge transfer spectra and not due to paramagnetism or d-d transition.

harge transfer complex involves electron transfer between metal ion and the ligands.
Hence, the statement is true.

Give a structure of $$Cr_2O^{2-}_7$$ ion.

1129811_168933_ans_ddddc9b0fca240c99d4ea2d8fefb33c1.png

Angle $$\alpha $$ in degrees is:

As we can see from the figure given that angle $$\alpha$$ is $$126^o$$.

Magnetic moment of $$Fe^{3+}$$ ions $$\sqrt {35}$$ BM. Determine number of unpaired electrons.

Magnetic moment $$=\sqrt{n(n+2)}$$

where $$n$$ is no. of electrons.

$$\sqrt{35}=\sqrt{n(n+2)}$$

$$\implies n=5$$ unpaired electrons.

Titanium is much harder than potassium or calcium and so, it is used in making jet aircraft engines.
If true enter 1, else enter 0.

Due to sharing of s as will as d-electrons, there is formation of stronger metallic bonding in Titanium and so, it is used in making jet aircraft engines.

Which of the following is not physically change ?
Magnetizing of a piece of iron.

A physical change is a temporary and reversible change in which the physical properties of the substance change without altering the composition of the substance. So, this is a physical change as no new substance is formed during the process .

No. of peroxide bonds in blue perchromate is two.
If true enter 1, else enter 0.

No. of peroxide bonds in blue perchromate is two.

What do you observe when water droplets fall on blue cobalt chloride ?

Blue cobalt chloride paper turns pink in the presence of water. Cobalt chloride in its hydrated form (having water) has color is pink.

$$CoCl_2 + 6H_2O \rightarrow CoCl_2.6H_2O$$

The transition metals (with the exception of Zn, Cd and Hg) are hard and have high melting and boiling points.
If true enter 1 else 0

Transition metals contain unpaired electrons. Hence, they form covalent bonds which results in hardness.

Transition metals have close packed structures in which transition metals are held together by strong metallic bonds which have notable covalent character.

Hence, transition metals have high melting and boiling points.

For Zn, Cd and Hg the d orbitals are completely filled and no covalent bonds are formed.

Transition elements show variable oxidation states . If true enter 1 else 0.

The energy difference between ns orbital and $$(n-1)d$$ orbital is small.
Hence, both ns and $$(n-1)d$$ electrons can be used for bond formation.
Hence, transition metals shows variable valence.

What is meant by disproportionation? Give two examples of disproportionation reaction in aqueous solution.

In disproportionation reactions, the same substance is oxidized as well as reduced.

For example,

(a) $$2H_2\underset{-1}{O_2} \rightarrow 2H_2\underset{-2}O + \underset{0}{O_2}$$

(b) $$\displaystyle 3\underset{+6}{Mn}O_4^{2-} + 4H^+ \rightarrow \underset{+7}{2Mn}O_4^- + \underset{+4}{Mn}O_2 + 2H_2O$$

Give one point of dissimilarity between lanthanides and actinides.

Lanthanides and actinides both are not radioacive. Except promethium, all lanthanides are non-radioactive. While all actinides are radioactive.

(a) Which metal in the first transition series (3d series) exhibits $$+ 1$$ oxidation state most frequently and why?
(b) Which of the following cations are coloured in aqueous solutions and why?
$$Sc^{3+}, V^{3+}, Ti^{4+},Mn^{2+}$$
(At. Nos. $$Sc = 21,\ V = 23,\ Ti = 22,\ Mn = 25$$)

$$(a)$$ $$Cu$$ exhibits $$+1$$ oxidation state most frequently as it electronic configuration is $$[Ar]3d^{10}4s^1$$ and of $$Cu^+$$ is $$[Ar]3d^{10}$$ which is stable pseudo-inert gas configuration that is why it shows $$+1$$ oxidation state.

$$(b)$$ These ions which have unpaired electrons shows colour in aqueous solution:

$$Sc^{+3}=0$$ unpaired electron

$$V^{3+}=2$$ unpaired electron

$$Ti^{+4}=0$$ unpaired electron

$$Mn^{+2}=2$$ unpaired electron

(a) How would you account for the following:
(i) The chemistry of actinoids is more complicated as compared to lanthanoids
(ii) Transition metals form complex compounds
(b) Complete the following equation:
$$2{Mn{ O }_{ 4 }}^{ - }+6{ H }^{ + }+S{{ O }_{ 3 }}^{ 2- }\longrightarrow $$

(a) (i) Chemistry of actinoids is more complicated than actinoids because

1) actinoid elements show a variety of oxidation states.

2) they are mostly radioactive (thus, studying their properties) becomes difficult.

3) they are mostly present in small quantities.

4) their half-life is extremely small.

(ii) Transition metals ions form coordination complexes because they have empty valance shell orbitals that can accept pairs of electrons from a Lewis base. Ligands must, therefore, be Lewis bases.

(b) $${ 2MnO }_{ 4 }^{ - }+6{ H }^{ + }+5{ O }_{ 3 }^{ 2- }\longrightarrow 2{ Mn }^{ +2 }+5{ SO }_{ 4 }^{ 2- }+3{ H }_{ 2 }O$$

What type of magnetism is shown by a substance if magnetic moments of domains are arranged in same direction?

The substance shows ferromagnetism as their magnetic moments are added to each other if the individual magnetic moments of domains of the atoms are aligned with one another and they point in the same direction.

(i) Name the elements of $$3d$$ transition series which shows maximum number of oxidation states. Why does it show so?
(ii) Which transition metal of $$3d$$ series has positive $${ E }^{ o }_{\left( { { M }^{ 2+ } }/{ M } \right)} $$ value and why?
(iii) Out of $${Cr}^{3+}$$ and $${Mn}^{3+}$$, which is a stronger oxidizing agent and why?
(iv) Name a member of the Lanthanoid series which is well known to exhibit $$+2$$ oxidation state.
(v) Complete the following equation
$$Mn{ O }_{ 4 }^{ - }+8{ H }^{ + }+5{ e }^{ - }\longrightarrow $$

(i) Manganese ($$3d^5 \ 4s^2$$) shows maximum number of oxidation state as its atoms have unpaired electrons in $$3d$$ orbitals. It shows all the oxidation state from +2 to +7.

(ii) $$Cu$$ has positive $$E^o_{(M^{2+} / M)}$$value, as high energy ($$\Delta_a + \Delta_iH$$) to transform $$Cu(s)$$ to $$Cu^{2+}$$(aq) is not balanced by hydration enthalpy.

(iii) $$Mn^{3+}$$ is stronger oxidising agent as the charge from $$Mn^{3+}$$ to $$Mn^{2+}$$ results in half filled, $$d^5$$configuration which has extra stability.

(iv) Europium, ($$Eu^{2+}$$) is formed by losing the two $$s$$ electrons and its ($$Xe)f^7$$ configuration acounts its formation.

(v) $$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$$

What are the transition elements ? Write two characteristics of the transition elements.

Elements which in their ground state or in any of their oxidation state have partially filled d-orbital are called transition elements. The name transition given to the elements of d-block is only because of their position between s-block and p-block elements.

The two characteristics of transition elements are:

(i) They show variable oxidation states.

(ii) They generally form coloured compounds.

(i) Why do transition elements show variable oxidation states?
(ii) Name the element showing maximum number of oxidation states among the first series of transition metals from $$Sc(Z = 21)$$ to $$Zn (Z = 30)$$.
(iii) Name the element which shows only $$+3$$ oxidation state.
(iv) What is lanthanoid contraction? Name an important alloy which contains some of the lanthanoid metals.

1.Transition elements show variable state oxidation in their compounds because there is a very small energy difference in between (n-1)d and ns orbitals. As a result, electrons of (n-1)d orbitals as well as ns-orbitals take part in bond formation.

2. Mn(Z-25) has the maximum number of unpaired electrons present in the d-subshell so it shows maximum oxidation states(+7)

3. Sc only exhibits a +3 oxidation state in these series.

4.The lanthanide contraction, i.e. the reduction in size of the $$Ln^{3+}$$ ion from $$La^{3+}$$(103 pm) to $$Lu^{3+}$$(86.1 pm), is often explained by the poor shielding of the 5s and 5p electrons by the 4f electrons.An alloy is a solid solution of two or more elements in a metallic matrix. It can either be a partial solid solution or a complete solid solution. Alloys are usually found to possess different physical properties than those of the component elements.An important alloy of lanthanoids is Mischmetal. It contains lanthanoids (94−95%), iron (5%), and traces of S, C, Si, Ca, and Al.

$$Ti^{4+}$$ ion is colourless. Give reason.

The electronic configuration of Ti is $$\displaystyle [Ar] 3d^24s^2$$
The electronic configuration of Ti$$\displaystyle ^{4+}$$ is $$\displaystyle [Ar] 3d^04s^0$$.
Since no unpaired electron is present in Ti$$\displaystyle ^{4+}$$, it is colourless.

$$CuI_2$$: Brown :: $$AgCl$$ : ________

$$CuI_2$$ shows brown colour in its natural state.

SImilarlu, silver chloride $$AgCl$$ occurs white in colour in nature.

Fill in the blanks by choosing the appropriate word/words from those given in the brackets:
(increases, decreases, positive, efficient, 68, non-efficient, no $$\alpha$$ -hydrogen, $$\alpha$$-hydrogen, negative, Rosenmund's, greater, Cannizzaro, 74, common-ion effect, lesser, buffer action, diamagnetic, paramagnetic)
The transition metals show ............. character because of the presence of unpaired electrons and $${Cu}^{+}$$ is ................... because its electronic configuration is $$[Ar]3{d}^{10}$$.

paramagnetic, diamagnetic

Give the equations for the conversion of argentite $$(Ag_2S)$$ to metallic silver.

$$\underset{argentite}{Ag_2S} + 4NaCl \rightleftharpoons 2Na[Ag(CN)_2] + Na_2S$$
$$Na_2S + 2O_2 \rightleftharpoons Na_2SO_4$$
$$2Na[Ag(CN)_2] + Zn \rightarrow Na_2 [Zn(CN)_4] + 2 Ag \underset{Silver \,metal}{\downarrow}$$

Why is $$S{ c }^{ 3+ }$$ colourless while $${ Ti }^{ 3+ }$$ coloured ? (Atomic number $$Sc=21, Ti=22$$)

Electronic configuration of $$_{ 21 }^{ }{ { Sc }^{ 3+ } }$$ is $$\left[ Ar \right]$$ $$3{ d }^{ 0 }$$ and $$ _{ 22 }^{ }{ { Ti }^{ 3+ } }$$ is $$\left[ Ar \right]$$ $$3{ d }^{ 1 }$$.
Therefore, due to absence of unpaired electron $${ Sc }^{ 3+ }$$ is colourless and due to presence of one unpaired electron $$d-d$$ transition takes hence $${ Ti }^{ 3+ }$$ is coloured.

Explain the following:
$$(i)$$ Why do transition metal ions possess a great tendency to form complexes?
$$(ii)$$ The paramagnetic character in $$3d$$-transition series elements increases upto $$Mn$$ and then decreases.

(i) The transition metal ions possess a great tendency to form complexes because of their small size, high ionic charge and availability of partially filled d-orbitals for bond formation. These partially filled d-orbitals can easily accomodate ligands electrons and consequently transition metals ions form complexes. Examples $${ \left[ Fe{ \left( CN \right) }_{ 6 } \right] }^{ 3- },Ni{ \left( CO \right) }_{ 4 }$$ etc.
(ii) In the $$3d$$ transition series as we move from $$Sc$$(21) to $$Mn$$(25), the number unpaired electrons increases and hence paramagnetic character increases. After $$Mn$$, the pairing of electrons in the d-orbital starts and the number of unpaired electrons decreases and hence, paramagnetic character decreases.

What is "Spitting of Silver"? How is it prevented?

Molten silver absorbs twenty times its volume of oxygen which it again expels on cooling. Globules of molten silver are thrown off. This is called spitting of silver. This can be prevented by covering the molten metal with a layer of charcoal.

Write a note of Plumbosolvency.

Lead is not attacked by pure water in the absence of air, but water is containing dissolved air has solvent action on lead due to the formation of lead hydroxide, a poisonous compound. This phenomenon is called Plumbo solvency.
$$2Pb+{ O }_{ 2 }+2{ H }_{ 2 }O\longrightarrow \underset { Lead\quad Hydroxide }{ 2Pb{ (OH) }_{ 2 } } $$

How does gold react with aquaregia ?

Gold dissolves in aquaregia $$( 3$$ parts of con$$.$$ $$HCl+1$$ part of con$$.$$ $$HNO_{3})$$ to form auric chloride$$.$$
$$2Au+9HCl+3HNO_{3}\rightarrow 2AuCl_{3}+6H_{2}O+3NOCl$$
auric chloride

Explain chromyl chloride test with equation.

When salt containing chloride is treated with $$K_2Cr_2O_7$$ and con. $$H_2SO_4$$ reddish orange vapours of chromyl chloride are obtained.
$$K_2Cr_2O_7 + 4KCl +6H_2SO_4 \rightarrow \underset{Chromyl \,\,chloride}{2CrO_2Cl_2 +6KHSO_4} +3H_2O$$
This reaction is used in the detection of chloride ions in qualitative analysis.

Why do d-block elements from complexes?

Transition elements have strong tendency to form complexes because of two reasons. They are,
i) Small size and high positive charge density.
ii) Presence of vacant$$(n-1)d$$ orbitals which are of appropriate energy of accept lone pair and unshared pair of electrons from the ligands for bonding with them.

Aqueous $$Cu^{+2}$$ ions are blue in color, where as aqueous $$Zn^{+2}$$ ions are colorless. Why?

The valence shell electronic configuration of $$\displaystyle Cu^{+2} $$ ion is $$\displaystyle 3d^9$$. It contains 1 unpaired electron.
The valence shell electronic configuration of $$\displaystyle Zn^{+2} $$ ion is $$\displaystyle 3d^{10}$$. It contains 0 unpaired electrons.
Hence, aqueous $$\displaystyle Cu^{+2} $$ ions are blue in color, where as aqueous $$\displaystyle Zn^{+2} $$ ions are colorless.
Note: Transition metal ions are coloured due to presence of unpaired electrons.

What is tailing of Mercury?

Mercury reacts with ozone to form an oxide. Meniscus of mercury is lost and mercurous oxide sticks to the glass. It is known as tailing of mercury.
On shaking with water, the meniscus is restored.

Give the composition of the following alloys:
(a) German silver
(b) Brass

(a) German silver is an alloy of co[[er, zinc and nickel. Its composition is copper, from 50% to 61.6%; zinc, from 19% to 17.2%; nickel, from 30% to 21.1%.
(b) Brass is an alloy of copper and zinc. Different types of brass have different proportions of zinc and copper.
(i) Alpha brasses : Cu > 65 % and Zn < 35 %
(ii) Alpha-beta brasses : Cu 55- 65 % and Zn 35-45 %
(iii) Beta brasses : Cu 50- 55 % and Zn 45-50 %
(iv) Gamma brasses: Cu 61- 67 % and Zn 33-39 %
(v) White brass : Cu < 50 % and Zn > 50 %

Write the characteristic properties of transition elements.

The characteristic properties of Transition elements:
(i) They show metallic properties such as malleability, ductility, metallic luster.
(ii) They have strong metallic bonds and compact structure. They have high density and high melting and boiling points.
(iii) They are good electrical and thermal conductors.
(iv) They show variable oxidation state.
(v) They form large number of complexes.
(vi) They have catalytic properties.
(vii) They form organometallic compounds.

What is spitting of silver? How can it be prevented?

Molten solver absorbs about twenty times its volume of oxygen which it again expels on cooling. Globules of molten silver are thrown off. This is called "spitting of silver". This can be prevented by covering the molten metal with a layer of charcoal.

Give the composition and uses of Nichrome.

Nichrome is an alloy of nickel and chromium, with some amount of iron and/or other elements. usually, 80% $$Ni$$ and 20% $$Cr$$ is present.
Nichrome is used in resistance wire. It is used in a very wide variety of devices where electric heating is required. It is used in some dental restorations (fillings). It is used in the explosives and fireworks industry as a bridgewire in electric ignition systems, such as electric matches and model rocket igniters.

What is "Tailing of Mercury"

Mercury reacts with ozone to form an oxide. Meniscus of mercury is lost and mercurous oxide sticks to the glass. It is known as tailing of mercury. On shaking with water, the meniscus is restored.

Write the uses of Lanthanides.

Lanthanides:
$$1$$. A pyrophoric alloy which contains cerium, lanthanum and Neodymium, iron; aluminium; calcium, carbon and silicon is used in cigarette lighters, toys, flame throwing tanks and tracer bullets.
$$2$$. Ceria $$(CeO_2)$$ and thoria $$(ThO_2)$$ are used in gas lamp materials.
$$3$$. Cerium salts are used in dyeing cotton, lead storage batteries and as catalyst.
$$4$$. Lanthanides are used in metallothermic reactions due to their extraordinary reducing property. Lanthanido-thermic processes can yield sufficiently pure Nb, Zr, Fe, Co, Ni, Mn, Y, W, U, B and Si.
$$5$$. Alloys of Lanthanides are known as mish-metals. The major constituents of mish-metals are $$Ce(45-50\%)$$, $$La(25\%), Nd(5\%)$$ and small quantities of other lanthanide metals and Fe and Ca impurities. Mish-metals are used for the production of brands of steel like heat resistant, stainless and instrumental steels. Mg-alloys containing $$30\%$$ mishmetal and $$1\%$$ Zr are useful in making parts of jet engines.

Write the equations involved in the preparation of potassium dichromate from chromite ore.

The balanced equations in the manufacture of $$\displaystyle K_2Cr_2O_7$$ from chromite ore are as given below:

Step I : roasting- conversion of chromite ore into sodium chromate
$$\displaystyle 4FeO \cdot Cr_2O_3 + 8Na_2CO_3+7O_2 \rightarrow 2Fe_2O_3 + 8Na_2CrO_4+8CO_2 $$

Step II : Conversion of sodium chromate into sodium dichromate

$$\displaystyle 2Na_2CrO_4+H_2SO_4 \rightarrow Na_2Cr_2O_7+Na_2SO_4+H_2O$$

Step III : Conversion of sodium dichromate into potassium dichromate

$$\displaystyle Na_2Cr_2O_7+2KCl \rightarrow K_2Cr_2O_7+2NaCl$$

Explain:
(i) Transition elements mostly form the complex compound. Why?
(ii) Transition elements are good catalyst. Why?
(iii) Transition elements show variable valency. Why?

(i) Tendency of transition element to form complexes compound is due to (1) Variable valency and different oxidation states. (2) Small size of transition elementions and high nuclear charge. (3) Availability of vacant d orbitals to accommodate lone pair of electrons donated by lingonds.
(ii) Transitional elements act as good catalysts in chemical reaction. In the hydrogenation Ni metal, In contact process of manufacture of SO, Pt and in manufacture of NH$$_3$$ by Haber process Fe acts as catalyst. In the method of preparation O$$_2$$ by heating KClO$$_3$$, MnO$$_2$$ acts as catalyst.
(iii) Transition metals exhibit variable valencies because the energy subshell (n - 1)d and ns are very close. Thus, possibility to lose electrons from ns subshell as well as from (n - 1)d subshell is very much if there an unpaired electrons. So oxidation states of these metals may increase.
In these elements Mn shows maximum variable valencies.

Write the factors which are related to the colour of transition metal ions.

Transition metal ions are coloured due to presence of unpaired electrons which results in d-d transitions of these unpaired electrons.

Transition metal ions have incomplete $$\displaystyle (n-1)$$ d orbitals. A small amount of energy can promote one or more electrons from a lower energy level to higher energy level within d orbitals with same principal quantum number.

This small amount of energy comes in visible region. Transition metal ion absorbs energy from visible region and appears coloured.

Explain the following
(i) $$Cu^+$$ is colourless and $$cu^{2+}$$ is coloured,
(ii) $$Zn$$ shows only + 2 Oxidation state.

(1) $$Cu^+$$ is colourless and $$cu^{2+}$$ is coloured : The colour of transition metal ions or their compound due to the partially filled $$(n -1)d$$ orbitals, i.e. the unpaired electrons in $$(n-1) d$$ orbitals. When white light ta on such a Metal compound some light energy corresponding to certain colouris absorbed and elections are excited to the set of higher energy. This type of promotion is called $$d-d$$ transition. The other colour of white light are transmitted which provide complementary coloursand compound appears coloured.
The transition metal ions which have completely filled orbitals or do not possess unpaired electron are colourless because there is no probability of $$d-d$$ transition.
(ii) $$Zn$$ shows only $$+ 2$$ oxidation state : $$Zn$$ represent only $$+2$$ Oxidationstate because due to participation of $$ns$$ and $$(n - 1) d$$ electrons inbonding, all transition elements show different oxidation states and form various compounds. Minimum oxidation state of these elements is $$+2$$ which is due to involvement of electrons of outer most orbit. Higher oxidation states of these elements are due to participation of d electrons of penultmateshall along with $$ns$$ electrons because the $$s$$ orbital of outer msot shell and $$d$$ orbital of penultimateshell arevery nearer to each other on energy scale. Therefore when some energy is supplied, electrons from $$(n-1) d$$ subshell are also removed and different oxidation are found.

With reference to the first row transition series :
i) Name the metal which possesses maximum number of oxidation states.
ii) Among $$Zn^{+2}$$ and $$Cu^{+2}$$ which is colourless?
iii) Between $$Ti^2+$$ and $$V^{2+}$$ which ion contains more number of unpaired electrons?

i) The metal which possesses maximum number of oxidation states are Cr (+1,+2,+3,+4,+5,+6) and Mn(+2,+3,+4,+5,+6,+7)

ii) $$\displaystyle Zn^{2+}$$ with 0 unpaired electrons is colourless. $$\displaystyle Cu^{2+}$$ with 1 unpaired electron is blue.

iii) $$\displaystyle Ti^{2+}$$ has 2 unpaired electrons. $$\displaystyle V^{2+}$$ has 3 unpaired electrons.
Hence, $$\displaystyle V^{2+}$$ on contains more number of unpaired electrons.

(a) What are Lanthanide elements?
(b) What do you understand by Lanthanide contraction?
(c) Write chromyl chloride test with equation.

(a) Lanthanide elements: Lanthanides are elements which resemble very much in properties with lanthanum. Lanthanide is group of $$14$$ elements from atomic number $$58$$ to $$71$$. In these elements on increasing atomic number electron enters into $$f$$-sub-shell of the atom. This sub-shell belongs to $$\left(n-2\right)$$ i.e., antipenultimate shell hence these elements are also called $$f$$-block elemetns.
(b) Lanthanide contraction: Decrease in atomic or ionic radii with increase in atomic number of lanthanides is called lanthanide contraction.
(c) Chromyl Test: Chloride radical is detected by this test. In chromyl chloride test, chromyl chloride gas (orange red colour is produced on heating the salt with solid $${K}_{2}{Cr}_{2}{O}_{7}$$ and conc. $${H}_{2}S{O}_{4}$$.
$$4NaCl + {K}_{2}{Cr}_{2}{O}_{7} + 6{H}_{2}S{O}_{4} \rightarrow 4NaHS{O}_{4} + 2KHS{O}_{4} + 3{H}_{2}O + 2Cr{O}_{2}{Cl}_{2}$$
(Orange-red gas)

Give any two differences between lanthanoids and actinoids.

Lanthanoids	Actinoids
Differentiating electron enters in 4f orbitals.	Differentiating electron enters in 5f orbitals.
Except promethium all elements occur in nature.	Except uranium and thorium all other elements are synthesized in laboratory. uranium and thorium occur in nature.

(a) Why Transition elements become paramagnetic?
(b) Explain oxidization properties of potassium permanganate in acidic medium. (any four only)

(a) In the atoms of paramagnetic elements, there should be more or less one lone pair of electrons. In the configuration of transition elements and its ions are found lone pair of electrons generally. Hence transition elements exhibit paramagnetic property.
(b) Oxidization properties of $$KMn{O}_{4}$$ in acidic medium:
In the acidic medium $$KMn{O}_{4}$$ solution $$Mn{O}^{-}_{4}$$ gives magnous ions $$\left({Mn}^{2+}\right)$$.
$$2Mn{O}^{-}_{4} + 6{H}^{+} \rightarrow 2{Mn}^{2+} + 3{H}_{2}O + 5\left[O\right]$$
Example: from $$KI$$, produces iodine
(i) $$2KMn{O}_{4} + 8{H}_{2}S{O}_{4} + 10KI \xrightarrow [ ]{ \quad Acidic } 6{K}_{2}S{O}_{4} + 2MnS{O}_{4} + 8{H}_{2}O + 5{I}_{2}$$
(ii) $$2KMn{O}_{4} + 5S{O}_{2} + 2{H}_{2}O \xrightarrow [ ]{ \quad Acidic } 2MnS{O}_{4} + 2{H}_{2}S{O}_{4} + {K}_{2}S{O}_{4}$$
(iii) $$2KMn{O}_{4} + 16HCl \rightarrow 2KCl + 2Mn{Cl}_{2} + 8{H}_{2}O + 5{Cl}_{2}$$
(iv) $$5\begin{matrix} COOH \\ | \\ COOH \end{matrix} + \left[5O\right] \rightarrow 10C{O}_{2} + 5{H}_{2}O$$

Calculate the value of magnetic moment of $$V^{+2}$$ ?

$$ \displaystyle V^{2+}$$ has outer electronic configuration of $$ \displaystyle 3d^3$$ with 3 unpaired electrons.
The magnetic moment $$ \displaystyle \mu _{eff} = \sqrt {n(n+2)}$$ B.M.
$$ \displaystyle \mu _{eff} = \sqrt {3(3+2)}$$ B.M.
$$ \displaystyle \mu _{eff} = \sqrt {15}$$ B.M.
$$ \displaystyle \mu _{eff} = 3.87$$ B.M.

Calculate the 'spin only' magnetic moment of $$M^{2+}(aq)$$ ion (Z=27).

Given ion is $$M^{2+}$$ with the atomic no. 27.

So, electronic configuration of $$M^{2+}$$ is [Ar] $$3d^7 4s^0$$

[Ar] $$\upharpoonleft $$$$\downharpoonright$$ $$\upharpoonleft $$ $$\downharpoonright$$ $$\upharpoonleft $$ $$\upharpoonleft $$ $$\upharpoonleft $$

3d 4s

No. of unpaired $$e^-$$ s =3

So, the spin only magnetic moment is given by,

$$\mu$$=$$\sqrt{n(n+2)}$$ =$$\sqrt{3(3+2)}$$ =$$\sqrt{3 X5}$$ =$$\sqrt{15}$$ BM

=3.87 BM

What is lanthanide contraction? Explain the cause of lanthanide contraction.

Lanthanide contraction is the gradual decrease in the atomic and ionic size of lanthanoids with an increase in atomic number.

Causes of lanthanide contraction:

With an increase in the atomic number, the positive charge on nucleus increases by one unit and one more electron enters same 4f subshell.

The electrons in 4f subshell imperfectly shield each other. Shielding in a 4f subshell is lesser than in d subshell.

With the increase in nuclear charge, the valence shell is pulled slightly towards the nucleus. This causes lanthanide contraction.

(i) Scandium (z = 21) is a transition element but zink (z = 30) is not. Explain.
(ii) Calculate the equivalent weight of $$KMnO_4$$in acidic medium.
(iii) What do you mean by Lanthanoid contraction ?

(i) On the basis of incompletely filled 3d orbitals, in case of $$Sc$$ atom in it's ground state $$\left( { 3d }^{ 1 } \right) $$, it is regarded as transition element. On the other hand, $$Zn$$ atom has completely filled d orbitals $$\left( { 3d }^{ 10 } \right) $$ in it's ground state as well as in this oxidised state, Hence, not regarded as transition element.

(ii) The molar mass of $${ KMnO }_{ 4 }$$ is 158 g/mol. To find equivalent mass, divide that molar mass by mole of electrons taken in it's half $$rxn$$. That would be $$5.\dfrac { 158 }{ 5 } =31.6$$ gms/equivalent.

(iii) Lanthnoid contraction is steady decreases in ionic radii of the elements in the lanthandie series from atomic number 57, lanthanum (to) 71, lutetium.

Out of $$C{ r }^{ 3+ }$$ and $$M{ n }^{ 3+ }$$, which is stronger oxidising agent and why?

The electronic configuration of Cr$$^{3+}$$ and Mn $$^{3+}$$

Cr$$^{3+}$$- [Ar}3d$$^3 4s^0$$.

Mn$$^{3+}$$ - [Ar}3d$$^4 4s^0$$.

So, when Cr$$^{3+}$$ and Mn$$^{3+}$$ act as oxidising agent, they get reduced to Cr$$^{2+}$$ and Mn$$^{2+}$$ respectively. The electronic configuration attained are 3d$$^4$$ and 3d$$^5$$ respectively.

Since Mn$$^{2+}$$ has half filled stable configuration so it is a stronger oxidising agent.

Why do transition elements form alloys?

Transition metals have very similar atomic sizes. One metal can easily replace the other metal from its lattice to form solid solution (alloy). Transition metals are miscible with one another in the molten state. The molten state solution of two or more transition metals on cooling forms alloy.

Explain why:
Zr and Hf exhibit similar properties.
(At. no. of Zr = 40, Hf = 72)

The transition metals of third series ($$Hf$$) have nearly the same radii as that of second transition series ($$Zr$$) due to the phenomenon called lanthanoid contractions which involves filling of f- orbitals. Because of the weak screening or shielding effect of f-orbital electrons in $$Hf$$ the effective nuclear charge on valance electrons is similar to that on $$Zr$$ and therefore results in similar properties.

What happens when sulphur dioxide gas is passed through an aqueous solution of a Fe(III) salt?

Since SO$$_2$$ is a reducing agent so, $$Fe^{3+} $$ is reduced to $$Fe^{2+} $$. The reaction is as follows:

$$2Fe^{3+} + SO_2+ 2H_2O$$ $$\rightarrow$$ $$2Fe^{2+}+SO_4^{2-}+4H^+$$

(i) Write down any three difference between Lanthanoid and Actinoid ?
(ii) The melting and boiling points of Zy. Cd and Hg are low. Why ?
(iii) Draw the structure of manganate ion.

(i)

A)All lanthanides are non radiaactive,but actines are radio active in the nature.

B)Lanthanides do not form oxo -cations but aidines form oxocatines such as $$UO^+,PuO^+,NPO_2^+$$.

C)Most of lanthanoids are colourless in nature but most of the actinides are coloured ions.

(ii)Melting and boiling points of $$Zn,Cd,Hg$$ are low.

(iii)Refer to image.

1044898_777411_ans_822e52d2f5684b5e8eb2c235047eb3ec.PNG

Explain why:
Transition elements usually form coloured ions.

The five degenerated d-orbital of transition elements in the presence of ligands splits into two sets of degenerated orbitals as $${ d }_{ { x }^{ 2 }-{ y }^{ 2 } },\ { d }_{ { z }^{ 2 } }\ and\ { d }_{ xy },{ d }_{ yz },{ d }_{ zx }$$. Thereby, upon absorption of light from visible region of the spectrum results in d-d electron transition of un-filled transition elements and reflects rest of the colours. This results in complimentary colour, thereby forms coloured ions.

What is meant by ferrimagnetism? Give any two examples.

A ferrimagnetic material is one that has populations of atoms with opposing magnetic moments. The opposing moments are unequal and a spontaneous magnetization remains. This happens when the populations consist of different materials or ions (such as $$Fe^{2+}$$ and $$Fe^{3+}$$).

Ferrimagnetism is exhibited by ferrites and magnetic garnets. Example: magnetite (iron(II,III) oxide; $$Fe_3O_4$$) and yttrium iron garnet are ferrimagnet.

Write any four applications of $$'f'$$ - block elements.

Applications of f-block elements:

1. Thorium is utilized as a part of cancer treatment and in glowing gas mantles.

2. Lanthanide oxides are utilized in cleaning glass as abrasives.

3. Uranium is utilized as an atomic fuel.

4. Plutonium is utilized as a part of nuclear reactors and in nuclear bombs

In the $$3$$d series from Sc$$(z=21)$$ to Zn$$(z=30)$$, enthalpy of atomisation of Zn is the lowest.

The Zn having the atomic number 30 and is an element of 3d series. The enthalpy of atomization depends on the extensive metallic bonding which in turn depends on the number of unpaired electrons. Due to the absence of these unpaired electrons in Zn. The interatomic bonding is weak in Zn and hence the enthalpy of atomization of Zn is the lowest.

What happens when $$(NH_4)_2Cr_2O_7$$ is heated?

(I) $$Zn+{ H }_{ 2 }{ SO }_{ 4 }\longrightarrow { ZnSO }_{ 4 }+{ H }_{ 2 }$$

(dil)

oxidizing agent $$\rightarrow \left( { H }^{ + } \right) $$

$$\because \quad { H }^{ + }$$ gains electrons

(II) $$Zn+{ H }_{ 2 }{ SO }_{ 4 }\longrightarrow { ZnSO }_{ 4 }+{ SO }_{ 2 }+{ H }_{ 2 }O$$

(conc)

oxidizing agent $$\rightarrow { SO }_{ 4 }^{ 2- }$$

$$\because $$ Sulphur gains electrons.

Magnetic momentum of an ion $${ M }^{ +3 }$$ is $$\sqrt { 35 }\ BM$$ Then, number of electron in $$d$$ orbital of $$m$$ element will be__________.

We know,

$$\mu=\sqrt{n(n+2)}$$

So,

$$\sqrt{35}=\sqrt{n(n+2)}$$

$$\implies n=5$$

So number of unpaired electrons in $$M^{3+}$$ ion is 5.

So number of electrons in orbital of d will be 6(If there is $$4s^2$$ configuration)

If not then number of electrons in d-orbital will be 9

Which of the following ions is coloured in aqueous solution?
i) $$Cu^{2+}$$
ii) $$Cu^{1+}$$

$$Cu^{2+}$$ is coloured because d-d transition. It is only possible when d-subshell have unpaired electron. $$Cu^{2+}$$ has partially filled orbital. Hence $$Cu^{2+}$$is coloured.
It is not colored. To be colored, you need some kind of electron transition within the atom/ion. A basic necessity would be either unpaired electrons or a possibility of resonance.The d orbitals are fully filled, which means no unpaired electrons. The absence of electron transitions cause $$Cu^+$$ to be a colorless ion.

Write the nomenclature and symbol of an element having atomic number 119?

The periodic table,has only $$118$$ elements till now been found,

The new elements with very high atomic numbers are so unstable that,only minute quanties,sometime only a few atoms of them oftained.

IUPAC has mentioned some rules for naming their.

According to the rules,

$$119=$$un-un-ennium

Symbol$$=$$uue

What is magnetic moment of $${ V }^{ +4 }$$?

Electronic configuration of $$V= [Ar] 3d^34s^2$$

Electronic configuration of $$V^{+4}=[Ar]3d^1$$

The number of unpaired electrons, n=1

Now we know,

$$\mu=\sqrt{n(n+2)}$$

put n=1,

$$\Rightarrow \mu=\sqrt{1(1+2)}=\sqrt3 BM$$

Magnetic moment of $$V^{+4}= \sqrt3BM$$

Why are metals like chromium and molybdenum very hard?

The d-orbitals containing unpaired electrons may overlap to form covalent bonds which are responsible for the hardness.
As the number of unpaired electrons increases, the number of covalent bonds and the strength of the metallic bonds increases. The increase in the number of covalent bonds results in an increase in the strength and hardness of the metal.
$$Cr$$ and $$Mo$$ have a maximum number of unpaired d-electrons which makes them very hard due to an increase in the number of covalent bonds.

Name the transition element of highest density.

Osmium is a transition element of highest density.

Density= 22.59g/cm³

Why basic character in lanthanoid hydroxides decreases across the period?

Basic character of the lanthanoid hydroxides

decreases from $$ Ce^{3+} $$ to $$ Lu^{3+} $$ because of

the decrease in the ions from $$ ce^{3+} $$ to $$ Lu^{3+} $$

The covalent character of M-OH bond increase

and hence the basic strength decreases

$$ \therefore Ce(OH)_{3} \equiv $$ most basic

$$ Lu(OH)_{3} \equiv $$ least basic

1114426_1146138_ans_e385f410fe714a958ab7728db0bc60fb.jpg

Transition metals and their compounds are used as catalysts.
Give two reasons.

Transition metals and their compounds function as catalysts either because of their ability to change oxidation state or, in the case of the metals, to adsorb other substances on to their surface and activate them in the process. All this is explored in the main catalysis section.

Write a chemical equation for burning of lanthanoids ($$ln$$) in oxygen.

$$ 4Ln+3O_{2}\rightarrow 2Ln_{2}O_{3} $$

Exception :- ce from $$ ceO_{2} $$ instead of $$ ce_{2}O_{3} $$

1114425_1146063_ans_95e061c6d69c4692a1c00601614f432a.jpg

Use hund's rule to derive the electronic configuration of $$Ce^{+3}$$ ion, and calculate its magnetic moment on the basis of 'spin only' formula. ( For Ce, Z = 58)

Electronic configuration of $$Ce:$$

$$Ce: 1s^22s^22p^63s^23p^63d^{10}4s^24p^64d^{10}5s^25p^64f^15d^16s^2$$

$$\therefore$$ Electronic configuration of $$Ce^{+3}$$ using hund's rule is,

$$Ce^{3+}: 1s^22s^22p^63s^23p^63d^{10}4s^24p^64d^{10}5s^25p^64f^1$$

or, $$Ce^{3+}: _{54}[Xe]4f^1$$

No. of unpaired electrons in $$Ce^{+3}=1=n$$

$$\therefore$$ Magnetic moment, $$\mu=\sqrt{n(n+2)}$$ $$B.M$$

$$\implies \mu=\sqrt{1(1+2)}$$ $$B.M$$

$$\implies \mu=\sqrt{3}$$ $$B.M$$

$$\implies \mu=1.73$$ $$B.M$$

Give any two difference between lanthanoids and actinoids.

Lanthanides are involved in the filling of 4f- orbitals whereas actinides are involved in the filling of 5f-orbitals.

The binding energy of 4f electrons is comparatively less than that of 5f-electrons.

Lanthanides are non-radioactive in nature except promethium whereas all actinide series elements are radioactive.

Answer the following question : Among the divalent cations in the first series of transition elements,manganese exhibits the maximum paramagnetism.

1352465_1280316_ans_5ed502ef9d7949d8b228103ac0e081af.jpg

Calculate the magnetic moment of $$MnS{O_4}$$?

1352473_1281258_ans_45a5d73218cb4f5b8f5e9cf47dc0d570.jpg

Transition elements generally forms coloured compounds. Why?

Transition elements have partially filled d orbitals. We also know that when electrons jump from one orbital to another light is emitted due to which the compounds of transition elements seem to be colored compounds.

Give two reactions in which transition metals or their compound acts as catalysts.

Reactions in which transition metals or their compound acts as catalysts-

Haeber's process $$\rightarrow$$ It is the formation of ammonia from nitrogen and hydrogen using iron as the catalyst. This is a heterogeneous system.

$${{N}_{2}}_{\left( g \right)} + 3 {{H}_{2}}_{\left( g \right)} \xrightarrow{Fe} 2 {N{H}_{3}}_{\left( g \right)}$$

Decomposition $$\rightarrow$$ Manganese dioxide catalyses the decomposition of hydrogen peroxide. This is a heterogeneous system.

$$2 {{H}_{2}{O}_{2}}_{\left( l \right)} \xrightarrow{Mn{O}_{2}} 2 {{H}_{2}O}_{\left( l \right)} + {{O}_{2}}_{\left( g \right)}$$

Why $$Zn^{2+}$$ is diamagnetic whereas $$Cr^{3+}$$ is paramagnetic ?

$${Zn}^{+2}$$ has the outermost electronic configuration $$3 {d}^{10}$$ and therefore it has no unpaired electron while $${Cr}^{+3}$$ has the outermost elctronic configuration as $$3{d}^{3}$$ and therefore, it has $$3$$ unpaired electron.

Hence $${Zn}^{+2}$$ is diamagnetic whereas $${Cr}^{+3}$$ is paramagnetic.

Write short note on Lanthanide contraction.

1359661_1472945_ans_61b30fc5b7aa46a68bf6a634f1fb4851.jpeg

Why do transition metals have high enthalpies of atomization?

Transition elements have high effective nuclear charge and a large number of valence electrons. Therefore, they form very strong metallic bonds which results in high enthalpies of atomisation.

How is potassium dichromate prepared from chrome iron ore?

Step 1 - Chromite are is fused with sodium or potassium carbonate in free excess of air

$$4FeCrO_4 + 8 Na_2 CO_3 + 7O_2 \to \underset{(yellow)}{8NO_2CrO_4} + 2FeO_3 + 8CO_2$$

Step 2 - Yellow solution of sodium chromate is filtered and acidified with sulfuric acid to give a solution from which orange sodium dichromate is obtained.

$$2Na_2CrO_4 + 2H^+ \to Na_2 Cr_2O_7 + 2Na + H_2O$$

$$\downarrow 2KCl$$

$$\underset{more soluble)}{K_2Cr_2O_7 + 2NaCl}$$

$$\therefore$$ sodium dichromate when fused with $$KCl$$ forms orange crystal of Potassium dichromate

Transition elements from complexes easily ? Justify.

Transition elements forms complexes easily due to the following reasons-

Small size and high nuclear charge.
Availability of vacant $$d$$-orbitals of suitable energy to accomodate lone pairs of electrons donated by the ligands.

Account for the following:
(i) Manganese shows maximum number of oxidation states in 3d series.
(ii) $$E^o$$ value for $$Mn^{3+} / Mn^{2+}$$ couple is much more positive than that for $$Cr^{3+} / Cr^{2+}$$.
(iii) $$Ti^{4+}$$ is colourless whereas $$V^{4+}$$ is coloured in an aqueous solution.

i) Because $$Mn \ 3d^5 4s^2$$ it has maximum no. of $$e^{\circleddash }s \ 5de^{\circleddash }S$$ & two $$5e^{\circleddash }s$$ to lose or share

So it can exhibits oxidation state of $$+2$$ to $$+7$$ which is maximum number

ii) $$\begin{matrix} Mn^{3+}&3d^4 \\ Mn^{2+} & 3d^5 \end{matrix}$$ as $$Mn^{3+} \to Mn^{2+}$$ more stable reduction is more feasible (Ref. image) more stable.

iii) $$\begin{matrix}Ti& 3d^24s^2 \\Ti^{4+} & 3d^04s^0\end{matrix}$$

no $$e^{\circleddash }s$$ in $$d$$ orbital so no $$d-d$$ transition.

$$\therefore$$ colourless. (Ref. image)

one unpaid $$e^{\circleddash } d-d$$ transition

$$\therefore V^{4+}$$ coloured.

1224813_1587748_ans_425901fe1fbf4a8a99bdc24e505aadc5.png

Identify the unknown compounds:

1545614_1722113_ans_995f705d98d3437fb559b692bffd6880.jpg

Fill in the blanks:
(N) Vegetable ghee is manufactured by the ..... of oils in presence of ..... as a catlayst.

hydrogenacion nikel.

Complete the following equations:

1544547_1721519_ans_60f0c5f886594fc3baa32f4074dcab22.jpg

Complete the following reactions:

1542937_1719512_ans_577e8541bb814fb0ad139763b826e537.jpg

Give reason :
Transition metals and their compounds show catalytic activities.

Transition metal and their compounds show catalytic properties because of the following reasons:
1. Due to their variable oxidation state, they form an intermediate complex, thereby providing a new path for reaction.
2. Availability of empty d-orbital which provides good surface area for reaction.

Write two similarities between chemistry of lanthanoids and actinoids.

The similarities between the chemistry of lanthanoids and actinoids are-
(i) The lanthanides actinides both belong to the category of inner transition metals.
(ii) They both contain $$14$$ elements.
(iii) Both show a contraction of radii.

(iv) Both shows mainly an oxidation state of +3 and both are electropositive and very reactive.

Write the preparation of the following :
$$Na_2 Cr_2 O_7$$ from $$Na_2CrO_4$$

(i) $$Na_2Cr_2O_7$$ can be prepared from $$Na_2CrO_4$$ in the following ways:-

The yellow solution of sodium Chromate $$(Na_2CrO_4)$$ is treated with sulphuric acid to give a solution from which $$Na_2Cr_2O_7.2H_2O$$ can be crystallized.

$$2Na_2CrO_4+2H^+\rightarrow Na_2Cr_2O_7+2Na^++H_2O$$
Yellow Orange

Complete the following reactions.

1542894_1719453_ans_7e0b23a428d348f280c2e0a061297d97.jpg

The above reaction gives two products. Write the structures of the products.

1610557_1722480_ans_78d8e72391e0481d933877a27ad49ea9.PNG

The minimum number of electrons having magnetic quantum number of zero for $$Ni$$ is :

$$Ni: 1s^2 2s^2 2p^6 3s^2 3p^6 3d^84s^2$$

Except 3d-subshell, remaining each subshell has two electrons whose magnetic quantum number is zero, and in 3d-subshell it has one electron.

Hence, total number of electrons $$= 2\times 6+1=13$$.

Match the compounds in column-I with their properties in column-II

(A) Oxidation state of Mn in $K_{2} M n O_{4}$ is +6 and is manufactured from pyrolusite ore.

$(B) K M n O_{4}$ acts as oxidising agent in acid, neutral and basic medium and is also manufactured from pyrolusite ore.

$(C) K_{2} C r_{2} O_{7}$ has chromium in +6 oxidation state. It acts as an oxidising agent in acidic medium and is obtained from the chromite ore.

$(D) K_{2} C r O_{4}$ also has chromium in +6 oxidation state. It can be obtained from the chromite ore.

Reactivity of transition elements decreases almost regularly from Sc to Cu.Explain.

Reactivity of transition elements decreases almost regularly from Sc to Cu.

The reason is:

Along a transition series, there is a gradual increase in the magnitude of ionisation enthalpy from left to the right. As the ionisation energy increases reactivity of the metals decreases and vice-versa.

Hence, from Sc to Cu ionisation energy gradually increases and thus reactivity decreases regularly.

The second and third rows of transition elements resemble each other much more than they resemble the first row. Explain why?

Due to lanthanoid contraction, the atomic radii of the second and third row transition elements is almost same. So, they resemble each other much more as compared to first row elements.

Although $$Zr$$ belongs to $$4d$$ and $$Hf$$ belongs to $$5d$$ transition series but it is quite difficult to separate them why ?

Due to lanthanoid contraction , they have almost same size (Zr, 160 pm) and (Hf , 159 pm)

It is difficult to separate Zr and Hf because they are similar in size. This enables them to have similar chemical properties, which makes their separation procedure very difficult.

$$NiCl_2$$ in the presence of dimethyl glycoxime (DMG) gives a complex which precipitates in the presence of $$NH_4OH$$, giving a bright red color.
Predict whether it is paramagnetic or diamagnetic.

$$\text{Given complex is }diamagnetic\text{ in nature. There is no unpaired electrons present as DMG is a strong ligand.}$$ $$\text{Thus, pairing is possible.}$$
1925918_1727686_ans_d87b3dae6f2442c780307ae472c22829.JPG

1925918_1727686_ans_d87b3dae6f2442c780307ae472c22829.JPG

Predict the spin only magnetic moment for:
$$Mn^{2+}$$

The electronic configuration of $${Mn}^{2+}$$ is [Ar]$${3d}^{5}$$$${4s}^{0}$$.

So, the number of unpaired electrons present in $${Mn}^{2+}$$ is 5.

The spin only magnetic moment is given by $$\mu$$ = $$\sqrt{n(n+2)}$$ where n= number of unpaired electrons.

The spin only magnetic moment for $${Mn}^{2+}$$ is $$\mu$$ = $$\sqrt{5(5+2)}$$

= $$\sqrt{35}$$

= 5.92 BM

Predict the spin only magnetic moment for:
$$Fe^{2+}$$

The electronic configuration of $${Fe}^{2+}$$ is [Ar]$${3d}^{6}$$$${4s}^{0}$$.

So, the number of unpaired electrons present in $${Fe}^{2+}$$ is 4.

The spin only magnetic moment is given by $$\mu$$ = $$\sqrt{n(n+2)}$$ where n= number of unpaired electrons.

The spin only magnetic moment for $${Fe}^{2+}$$ is $$\mu$$ = $$\sqrt{4(4+2)}$$

= $$\sqrt{24}$$

= 4.90 BM

The type of magnetism exhibited by $$[Mn(H_2O)_6]^{2+}$$ ion is ________.

The type of magnetism exhibited by $$[Mn(H_2O)_6]^{2+}$$ ion is paramagnetism. It has 5 unpaired electrons since water is not a strong field so pairing does not occur and has $$sp^3d^2$$ hybridisation.

Answer the following:
Why the compounds of transition elements are coloured?

Transition elements show coloured compunds. The compound in which metal ion has an unpaired electron that compounds are coloured. When an electron from a lower energy d-orbital is excited to a higher energy d-orbital,if the energy of excitation lies in the visible region , complementary colour is observed.

Explain the following giving proper reason:

Transition metals and their compounds act as catalysis.

1569168_1729768_ans_d52a5d5483844f028130953ae7be2153.jpg

Explain the following giving proper reason:

$$CuSO_4$$ is paramagnetic while $$ZnSO_4$$ is diamagnetic.

1680932_1729756_ans_95072a227c864dbc8c14918090cf6545.jpg

Predict the spin only magnetic moment for:
$$Cu^{+}$$

1568292_1729446_ans_2982153072314d1c8ed4424bfc0201f7.jpg

Explain by giving suitable reason.

Yellow coloured aqueous solution of sodium chromate changes to orange-red when $$CO_2$$ under pressure is passed.

1569328_1729785_ans_46292e71bf3848f99f400bf3284293f4.jpg

Predict the spin only magnetic moment for:
$$Cr^{6+}$$

1568279_1729444_ans_45640325359b462e9f76d7d95b9ef393.jpg

Calculate magnetic moments of $$M^{2+}$$ ions of $$Ti(Z=22), Cr(Z=24), Mn(Z=25)$$ and $$Cu(Z=29)$$ and arrange them in increasing order.

1569293_1729782_ans_0bbc573ec7484111a0938b77061944c9.jpg

Answer the following:
Why most of the compounds of transition elements are paramagnetic?

The substances which are attracted by the magnetic field are called paramagnetic. This phenomenon are called paramagnetism. This property is showed only when the substance contain one or more unpaired electrons.Most of the transition elements show paramagnetic behaviour.The unpaired electrons i the (n-1)d orbitals are responsible for the magnetic properties.

Explain by giving suitable reason.

$$Hg^{2+}$$ and $$Hg_2^{2+}$$ salts are colourless.

The salts of $$Hg^{2+}$$ and $$Hg^{2+}_{2}$$ are colourless.The electronic configuration of Hg is $$[Xe]4f^{14}5d^{10}6s^{2}$$.

In $$Hg^{2+}$$, the oxidation state is +2 which means it loses its two electrons and then it has completely filled d-orbitals and also in the case of $$Hg^{2+}_{2}$$ loses one electron. In this salts, d-d transition is not possible as it requires high energy and doesn't fall within the visible region.

Hence, they are colourless.

Answer the following:
Which bivalent ion of first transition series show the highest paramagnetism?

Manganese ($$Mn^{2+}$$) show highest paramagnetism in the first transition series due to the maximum number of unpaired electrons i.e. 5 which is maximum in transition metal ion.

The electronic configuration of $$Mn^{2+}$$ is [Ar]$${3d}^{5}$$$${4s}^{0}$$.The paramagnetic character of any element is due to the presence of an unpaired electron.

There are 5 unpaired electrons.

Explain:
Which divalent metal ion has maximum paramagnetic character among the first transition metals? Why?

$$Mn^{+2}$$ , a divalent metal ion of the first transition series shows maximum paramagnetic character because it has maximum number of unpaired electrons.

The electronic configuration of $$Mn^{+2}$$ is [Ar]$$3d^{5}$$ i.e 5 unpaired electrons which is maximum in the first transition series.

A black mineral $$(A)$$ on heating in presence of air gives a gas $$(B)$$.

1570275_1730038_ans_c33d3dc3c07f4e56a6d1fe7bf0c020dd.jpg

Explain by giving suitable reason.

$$Cu^{2+}$$ salts are paramagnetic while $$Cu^{+}$$ salts are diamagnetic in nature.

The paramagnetic nature of any compound is due to its unpaired electrons. Copper(Cu) i.e. $$[Ar]3d^{10}4s^{1}$$ attained the full-filled stable electronic configuration. The electrons from 4s subshell jumped to 3d subshell and becomes more stable.

The salts of $$Cu^{2+}$$ are paramagnetic in nature while $$Cu^{+}$$ salts are diamagnetic. The electronic configuration of $$Cu^{2+}$$ and $$Cu^{+}$$ are $$[Ar]3d^{9}4s^{0}$$ and $$[Ar]3d^{10}4s^{0}$$. As $$Cu^{2+}$$ contains one unpaired electron hence, it's salts show paramagnetic behaviour whereas $$Cu^{+}$$ contains no unpaired electron and therefore it shows diamagentic behavior.

Write down the name of catalyst for the following:
Reppe synthesis of benzene.

2039917_1732042_ans_b7672b22c10a4c12a327bbb5ca0a57d0.jpg

Write down the name of catalyst for the following:
Fenton's reagent in the synthesis of aldehydes from alcohols.

Fenton's reagent is a solution of hydrogen peroxide ($$H_2O_2$$) with ferrous ions. Here iron(ii) sulphate (i.e $$FeSO_4$$) act as catalyst to oxidise aldehydes to alcohols.

Write down the name of catalyst for the following:
Deacon's process of making $$Cl_2$$ from $$HCl$$.

Deacon's process involves oxidation of hydrogen gas by atmospheric oxygen in the presence of $$CuCl_2$$ (catalyst) at 723 K.

$$4HCl +O_2 \xrightarrow {CuCl_2} 2Cl_2 + 2H_2O$$

$$CuCl_2$$ is acting as catalyst.

Write down the name of catalyst for the following:
Adams catalyst in reduction.

1576239_1732039_ans_f632b908f0cb41d68671902e24cbc003.jpg

Calculate magnetic moment of $$Fe^{3+}$$ in $$[Fe(CN)_6]^{3-}$$ and in $$[Fe(H_2O)_6]^{3-}$$.

1576306_1732047_ans_78403cb9299c4e63a40902c700aaa8a9.jpg

Match the List $$I$$ with List $$II$$:

1713564_1731819_ans_b89e3ed6dfa643a8b9c916a86115feb9.jpg

Give the formula of three ions which are coloured due to charge transfer spectra.

1576208_1732049_ans_e8f5955f31ee4f5aa1782f34f8190ee1.jpg

What is the magnetic moment of $$Fe^{3+}$$?

1722250_1732057_ans_98245b956ca04a0e959d53df9712c797.jpg

Arrange following complex in decreasing order of magnetic moment:

$$[Ni(H_2O)_4]^{2+},\ [Ni(CN)_4]^{2-},\ [Fe(CN)_6]^{3-},\ [Fe(CN)_6]^{4-}$$

1722239_1732048_ans_efe802b5ef4f43d891ac95e673570b00.jpg

What happen when?
Ammonium dichromate is heated strongly.

1576211_1732052_ans_50579be5d7b647e4a9f3d509b051d855.jpg

How many Cr-O bond are equivalent in chromate dianion?

1916692_1748706_ans_623f87241f674c7ea3370e5db31d2a0c.jpg

Match the items under List 1 with the compounds in ListSelect the correct answer from the sets (a), (b), (c) and (d).

$$ Al_2O_3 $$ is Corrundum

$$Fe_2O_3$$ is Haemetite

$$HgS$$ is Cinnabar

$$PbS$$ is Galena

Write similarity and difference between the chemistry of lanthanoid and actinoid elements.

Similarities:
i. Both show mainly an oxidation state of $$3+$$.
ii. Actinoids show actinoid contraction like lanthanoid contraction is shown by lanthanoids.
iii. Both are electropositive and very reactive.

Differences:
i. Except promethium(Pm) lanthanoids are non-radioactive whereas actinoids are radioactive.
ii. Lanthanoids do not form oxocation whereas actinoids form oxocation.
iii. Lanthanoids have less tendency towards complex formation whereas actinoids have greater tendency towards complex formation.

Explain: The metal-metal bonding is more frequently found with the second and third series of transition elements.

In the same group of $$d$$-block elements, the $$4d$$ and $$5d$$ transition elements are larger in size than those of the corresponding $$3d$$ elements. Hence, the valence electrons are less tightly held and form metal-metal bond more frequently.

Which transition metal of $$3d$$ series has positive $$E^{\circ}(M^{2+}/M)$$ value and why?

Copper has $$E^{\circ}(M^{2+}/M)$$ value, as high energy ($$\Delta_{a}H ; \Delta_{i}H$$) to transform $$Cu_{(s)}$$ to $$Cu^{2+}_{(aq)}$$ is not balanced by hydration enthalpy.

Orange colour of $$Cr_{2}O_{7}^{2-}$$ ion changes to yellow when treated with an alkali. Why?

Orange colour of $$Cr_{2}O_{7}^{2-}$$ ion changes to yellow when treated with an alkali due to formation of chromate ion which is yellow in colour.

$$\underset{\overset{Dichromate\ ion}{(Orange)}}{Cr_{2}O_{7}^{2-}} + 2OH^{-}\rightarrow 2\underset{\overset{Chromate\ ion}{(Yellow)}}{CrO_{4}^{2-}} + H_{2}O$$

Account for: Transition metals form large number of complex compounds.

The tendency to form complex compounds is due to:
i. Small size and high charge on metal ion .
ii. The availability of $$d$$ orbitals for accommodating electrons donated by the ligand.

Explain: Transition elements and their compounds are known as catalysts.

The catalytic activity of transition metals and their compounds is attributed to the following reasons:

i. Due to their tendency to show variable oxidation states transition metals form unstable intermediate compounds and provide a new path for the reaction with lower activation energy.

ii. In some cases, the transition metals provide a suitable large surface area with free valencies on which reactants are adsorbed.

Chemistry of actinoids in complicated as compared to lanthanoids. Give two reasons.

Chemistry of actinoids in complicated as compared to lanthanoids because of the following reasons:
i. Actinoids are radioactive.
ii. Actinoids show wide range of oxidation states as energy difference between $$5f$$, $$6d$$ and $$7s$$ subshells of the actinoids is small.

Complete and balance the following reaction:

$$Cr_{2}O_{7}^{2-} + I^{-} + H^{+} \rightarrow$$

$$Cr_{2}O_{7}^{2-} + 14H^{+} + 6e^{-} \rightarrow 2Cr^{3+} + 7H_{2}O$$

$$ [2I^{-} \rightarrow I_{2} + 2e^{-}] \times 3$$
$$-----------------------$$
$$Cr_{2}O_{7}^{2-} + 6I^{-} + 14H^{+} \rightarrow 2Cr^{3+} + 3I_{2} + 7H_{2}O$$
$$-----------------------$$

Account for: $$Zr$$ and $$Hf$$ have almost similas atomic radii.

$$Zr$$ and $$Hf$$ have almost similar atomic radii due to filling of $$4f$$ orbitals which have poor shielding effect (Lanthanoid contraction).

What is lanthanoid contraction? Name an important alloy which contains some of the lanthanoid metals.

The steady decrease in the atomic and ionic radii (having the same charge) with increase in atomic number as we move across the series from lanthanum to lutetium is known as lanthanoid contraction.
Misch metal.

Account for: $$E_{0}$$ value for the $$Mn^{3+}/Mn^{2+}$$ couple is highly positive ($$+1.57V$$) as compare to $$Cr{3+}/Cr^{2+}$$ which is negative ($$-0.4V$$). Why?

The positive $$E_{0}$$ value for the couple $$Mn^{3+}/Mn^{2+}$$ is due to the much higher third ionisation energy of $$Mn$$ (where the required change is half filled $$d^{5}$$ (extra stable) to $$d^{4}$$. As $$Cr^{3+} (t^{3}_{2g})$$ is more stable than $$Cr^{2+}$$, therefore, $$E_{0}$$ value for couple $$Cr^{3+}/Cr^{2+}$$ is negative.

Account for: $$Zn$$ has lowest enthalpy of atomisation in $$3d$$ series.

Because in the formation of metallic bonds no electron from $$3d$$ orbitals are onvolved in case of zinc as it has completely filled $$d$$-orbitals ($$3d^{10}4s^{2}$$), while in all other metals of $$3d$$ series electrons from the $$d$$-orbitals are always involved in the formation of metallic bonds.

How do you prepare: $$Na_{2}Cr_{2}O_{7}$$ from $$Na_{2}CrO_{4}$$?

$$Na_{2}CrO_{4}$$ is extracted with water and acidified with $$H_{2}SO_{4}$$ to give $$Na_{2}Cr_{2}O_{7}$$

$$2\underset{Sodium\ chromate}{Na_{2}CrO_{4}} + H_{2}SO_{4} \rightarrow \underset{Sodium\ dichromate}{Na_{2}Cr_{2}O_{7}} + Na_{2}SO_{4} + H_{2}O$$

Account for: Transition metals form coloured compounds.

This is due to $$d-d$$ transition. Under the influence of the ligand attached, the $$d$$-orbitals in a transition metal complex split in two sets of orbitals having slightly different energies. When white light falls on these complexes they absorb a particular colour from the white light for the promotion of electrons and remaining colours are emitted. The colour of the complex is due to this emitted radiations.

Account for: $$Cr^{2+}$$ is a strong reducing agent.

$$Cr^{2+}$$ is a strong reducing agent because after the loss of one electron $$Cr^{2+}$$ becomes $$Cr^{3+}$$ which has more stable $$t^{3}_{2g}$$ (half-filled) configuration.

Define lanthanoid contraction. Write the common oxidation states of Lanthanoids

The overall decrease in atomic and ionic radii from lanthanum to lutetium, due to poor shielding effect of $$4f$$ electrons is called lanthanoid contraction. The common oxidation state fo lanthanoids is $$+3$$.

With reference to structural variability and chemical reactivity, write the differences between lanthanoids and actinoids.

Lanthanoids

i. Atomic or ionic radii does not show variation.

ii. Besides $$+3$$ oxidation state, they show $$+2$$ and $$+4$$ oxidation states in few cases.

Actinoids
i. Atomic or ionic radii shows much much variation.
ii. Besides +3 oxidation states, they show higher oxidation states of $$+4$$, $$+5$$, $$+6$$, $$+7$$ also.

Out of $$Mn^{3+}$$ and $$Cr^{3+}$$, which is more paramagnetic and why? (Atomic numbers: $$Mn=25$$, $$Cr=24$$)

$$Mn^{3+}(3d^{4}4s^{0})$$ has $$4$$ unpaired electrons while $$Cr^{3+}(3d^{3}4s^{0})$$ has $$3$$ unpaired electrons, therefore, $$Mn^{3+}$$ is more paramagnetic than $$Cr^{3+}$$.

Why are transition elements known as $$d-$$block elements?

The last electron enters $$(n-1)$$ $$d$$-orbital, i.e. $$d-$$ orbital of the penultimate shell. Hence, these are known as $$d-$$block elements.

Compare non transition and transition elements on the basis of their stability of oxidation states.

In transition elements higher oxidation states are favoured by heavier elements whereas in non transition elements lower oxidation state is favoured by heavier elements.

On the basis of lanthanoid contraction, explain the stability of the complexes of lanthanoids.

Stability of the complexes increases as the size of lanthanoids decreases.

Write the equations involved in the preparation of potassium dichromate from sodium chromate ($$Na_{2}CrO_{4}$$).

Sodium chromate is acidified with sulphuric acid to give a solution from which orange sodium dichromate, $$Na_{2}Cr_{2}O_{7}.2H_{2}O$$ can be crystallised.

$$2Na_{2}CrO_{4} + 2H^{+} \rightarrow Na_{2}Cr_{2}O_{7} + 2Na^{+} + H_{2}O$$

Sodium dichromate is more soluble than potassium dichromate. The latter is therefore prepared by treating the solution of sodium dichromate with potassium chloride.

$$Na_{2}Cr_{2}O_{7} + 2KCl \rightarrow K_{2}Cr_{2}O_{7} + 2NaCl$$

Give reason: From element to element, the actinoid contraction is greater than the lanthanoid contraction.

From element to element, the actinoid contraction is greater than the lanthanoid contraction due to poor shielding by $$5f$$-electrons in actinoids than that by $$4f$$-electrons in the lanthanoids.

When an oxide of manganese ($$A$$) is fused with KOH in the presence of an oxidising agent and dissolved in water, it gives a dark solution of compound ($$B$$). Compound ($$B$$) disproportionates in neutral or acidic solution to give purple compound ($$C$$). An alkaline solution of compound ($$C$$) oxidises potassium iodide solution to a compound ($$D$$) and compound ($$A$$) is also formed. Identify compounds $$A$$ to $$D$$ and also explain the reactions involved.

$$\underset{(A)}{2MnO_{2}} + 4KOH + O_{2} \rightarrow \underset{(B)}{2K_{2}MnO_{4}} + 2H_{2}O$$

$$3MnO_{4^{2-}} + 4H^{+} \rightarrow \underset{(C)}{2MnO_{4}} + MnO_{2} + 2H_{2}O$$

$$2MnO_{4} +H_{2}O + KI \rightarrow \underset{(A)}{2MnO_{2}} + 2OH +\underset{(D)}{KIO_{3}}$$

The elements of $$3d$$ transition series are given as :
$$Sc,\ Ti,\ V,\ Cr,\ Mn,\ Fe,\ Co,\ Ni,\ Cu,\ Zn$$

Which element is soft and why?

$$Zn$$, in $$Zn (3d^{10}4s^{2})$$ all the electrons present in $$d$$-orbitals are paired and hence metallic bonds present in it are weak. That is why, it is soft.

Hence $$Cu,\ Zn$$ are soft metals.

Account for: The atomic radii of the metals of the third ($$5d$$) series of transition elements are virtually the same as those of the corresponding members of the second ($$4d$$) series.

The atomic radii of the metals of the third ($$5d$$) series of transition elements are virtually the same as those of the corresponding members of the second ($$4d$$) series due to filling of $$4f$$-orbitals which have poor shielding effect (lanthanide contraction).

Draw the structures of chromate and dichromate ions.

1641550_1778887_ans_86f7e169c90a4882b266e6c4d9131761.png

Why does a transition series contain 10 elements?

There are five $$d$$-orbitals in an energy level and each orbital can contain two($$2$$) electrons. As we move from one element to the next, an electron is added and for complete filling of the five($$5$$) $$d$$-orbitals, $$10$$ electrons are required.

Describe the trends in the Magnetic behaviour of dipositive gaseous ions ($$M^{2+}$$) of the first series of the transition elements:

Except $$Zn^{2+}$$, all other divalent gaseous ions of the first series of the transition elements contain unpaired electrons in their $$3d$$-subshell and are therefore paramagnetic in nature.

Explain: There is a close similarity in physical and chemical properties of the $$4d$$ and $$5d$$ series of the transition elements, much more than expected on the basis of usual family relationship.

This is because $$5d$$ and $$4d$$-series elements have virtually the same atomic and ionic radii due to lanthanide contraction. Due to equality in size of $$Zr$$ and $$Hf$$, $$Nb$$ and $$Ta$$, $$Mo$$ and $$W$$, etc., the two elements of each pair have the same properties.

How would you account for: Many of the transition elements and their compounds can act as good catalysts.

The catalytic activity of transition metals is attributed to the following reasons:
a. Because of their variable oxidation states transition metals form unstable intermediate compounds and provide a new path with lower activation energy for the reaction.
b. In some cases, the transition metal provides a suitable large surface area with free vacancies on which reactants are adsorbed.

Explain why $$[Fe(H_2O)_3]^{3+}$$ has magnetic moment value of $$5.92\ BM$$ whereas $$[Fe(CN)_6]^{3-}$$ has a value of only $$1.74\ BM$$.

$$[Fe(CN)_6]^{3-}$$ involves $$d^2sp^3$$ hybridisation with one unpaired electron and $$[Fe(H_2O)_6]^{3+}$$ involves $$sp^3d^2$$ hybridisation with five unpaired electrons.

This difference is due to the presence of strong ligand $$CN^-$$ and weak ligand $$H_2O$$ in these complexes.

How would you account for: Transition metals and their compounds show catalytic properties.

The catalytic activity of transition metals and their compounds is attributed to the following reasons:
i. Due to their tendency to show variable oxidation states transition metals form unstable intermediate compounds and provide a new path for the reaction with lower activation energy.
ii. In some cases, the transition metals provide a suitable large surface area with free valencies on which reactants are adsorbed.

How would you account for: There is a greater range of oxidation states among the actinoids than among the lanthanoids.

This is due to more comparable energies of $$5f$$, $$6d$$ and $$7s$$ orbitals in actinoids.

Name an ore used in the preparation of potassium dichromate.

Chromite ore ($$FeCr_{2}O_{4}$$)

What is the general formula by which the electronic configuration of the transition elements is represented?

$$(n- 1)d^{1- 10}ns^{1-2}$$

Giving the name, the stereochemistry and the magnetic behaviour of the following complex:
$$[Co(NH_3)_5 Cl]Cl_2$$

IUPAC name: Pentaaminechloridocobalt $$(III)$$ chloride
$$Co(III): [Ar]3d^6 4s^0-d^2 sp^3$$ hybridisation leads to octahedral shape.
Magnetic behaviour: Diamagnetic.

Answer the following question:
Why is $$[NiCl_4]^{2-}$$ paramagnetic while $$[Ni(CN)_4]^{2-}$$ is diamagnetic?
(Atomic number $$Ni=28$$)

In both $$[NiCl_4]^{2-}$$ and $$[Ni(CN)_4]^{2-}, Ni$$ is in $$+2$$ oxidation state with configuration $$3d^8 $$ and it contains two unpaired electrons. In $$[NiCl_4]^{2-}$$ due to presence of weak ligand $$Cl^-$$ no pairing takes place and hence it is paramagnetic whereas in $$[Ni(Cl)_4]^{2-}, CN^-$$ is a strong field ligand and pairing occurs and hence it becomes diamagnetic.

Name the type of elements , which have their
Two outermost shell incomplete - ...............

Two outermost shell incomplete - $$\textbf {Transition elements }$$

How would you account for the following?
$$[Fe (CN)_6]^{3-}$$ is weakly paramagnetic while $$[Fe(CN)_6]^{4-}$$ is diamagnetic.

Due to the presence of one unpaired electron in the $$3d$$ subshell in $$[Fe(CN)_6]^{3-}$$, it is weakly paramagnetic.

On the other hand, $$[Fe(CN)_6]^{4-}$$ is diamagnetic because all the electrons are paired.

Mention any three processes where transition metals act as catalysts.

$$N_2(g) +3H_2(g) \overset{\rightharpoonup}{\leftharpoondown} 2NH_3(g)$$
In the above reaction, $$Fe$$ is used as catalyst.
$$KClO_3 \overset{Heat}{\underset{MnO_2}{\longrightarrow}} 2 KCl + 3O_2$$
$$2SO_2 +O_2 \overset{V_2 O_5}{\longrightarrow}$$ 2 SO_3$$

Name the reagents used in the following reactions:
$$C_6H_5-CH_2-CH_3\ \to \ C_6H_5-COO^- K^+$$

The reagents used in this reaction are Alkaline potassium permanganate $$(KMnO4) \ and \ \ KOH $$

What is the role of metal or regents written on arrows in the given chemical reactions?

Ni acts as a catalyst in this reaction.

Transition metals can act as catalysts because these can change their oxidation state. How does $$Fe(III)$$ catalyse the reaction between iodide and persulphate ions?

The reaction between iodide and persulphate ions.
$$2I^- + S_2 O^{2-}_8 \rightarrow I_2 + 2 SO_4^{2-}$$
An explanation of this catalytic action can be given as:
$$2 Fe^{3+} + 2I^- \rightarrow 2Fe^{2+} +I_2$$
$$2 Fe^{2+} + S_2 O^{2-}_8 \rightarrow 2Fe^{3+} +2 SO_4^{2-}$$

Complete each synthesis by giving missing reagents or products in the following :

1645271_1780613_ans_f659dd82b18a499fabde7d67b72af668.png

What is the role of metal or reagents written on arrows in the given chemical reactions?
(b) $$CH_3COOH + CH_3CH_2OH \xrightarrow {conc. \ H_2SO_4} CH_3COOC_2H_5 + H_20$$
(c) $$CH_3CH_2OH \xrightarrow[Heat]{KMnO_4} CH_3COOH$$

(a) Catalyst
(b) Catalyst
(c) Oxidizing agent

Correct the statement.
Transition elements are placed at extreme right of the periodic table.

$$\textbf {Noble gases}$$ are placed at the extreme right of the periodic table.

Name two elements in each case :
Actinides

2 examples are : Uranium and Neptunium

Compare the chemistry of actinoids with that of the lanthanoids with special reference to:
Chemical activity.

In the lanthanoide series, the earlier members of the series are more reactive. They have reactivity that is comparable of $$Ca$$. With an increase in the atomic number, the lanthanides start behaving similar to $$Al$$.

Actinoids, on the other hand, are highly reactive metals, especially when they are finely divided. When they are added to boiling water, they give a mixture of oxide and hydride. Actinoids combine with most of the non metals at moderate temperatures. Alkalies have no action on these actinoids. In case of acids, they are slightly affected by nitric acid (because of the formation of protective oxide layer).

Compare the chemistry of actinoids with that of the lanthanoids with special reference to:
Atomic and ionic size.

Similar to lanthanoids, actinoids also exhibit actinoid contraction (overall decrease in atomic and ionic radii). The contraction is greater due to the poor shielding effect of 5f orbitals.

Compare the chemistry of actinoids with that of the lanthanoids with special reference to electronic configuration.

The general electronic configuration for lanthanoids is $${[Xe]}4f^{0-14}$$ $$5{d}^{0-1}6{s}^{2}$$ and that of actinoids is $$[Rn]5{f}^{0-14}6{d}^{0-1}7{s}^{2}$$.

Unlike $$4f$$ orbitals, $$5f$$ orbitals are not deeply buried and participitate in bonding to a greater extent.

Explain giving reasons:
The transition metals generally form coloured compounds.

Colour of transistion metal - compounds is due to the excitation of an electron from a lower energy d-orbital to a higher energy $$d$$ orbital. The energy of excitation corresponds to the frequency of light absorbed and the colour observed corresponds to the complementary color of the light absorbed (whose frequency lies on the visible region).

The frequency of the light absorbed depends on the nature of the ligand. Transition metals form colored compounds due to the presence of vacant d-orbitals from the d-d transition of $$electrons$$ which causes the color.

Explain giving reasons:
Transition metals and many of their compounds show paramagnetic behaviour.

Paramagnetic arises due to the presence of unpaired electrons. Each such electrons has a magnetic moment associated with it due to its spin angular momentum. Transition metals have in its ground state or ionised state have number of unpaired d-electrons which gives it a paramagnetic behaviour.

Write any three differences between lanthanides and actinides.

Differences between Lanthanides and Actinides:

1) Lanthanides involve the filling of $$4f$$ orbitals while actinides involve the filling of $$5f$$ orbitals in their atoms.

2) Energy differences between the $$5f, 6d$$ and $$7s$$ orbitals are relatively small, hence multiple oxidation states are possible for Actinides than those of the lanthanides.

3) Lanthanides have less tendency towards the complex formation. And except promethium, all lanthanides are non-radioactive, whereas actinides have a stronger tendency towards the complex formation and all the actinides are radioactive.

Compare the chemistry of actinoids with that of the lanthanoids with special reference to:
Oxidation states.

The principal oxidation states of lanthanoids is $$+3.$$ However sometimes we also encounter oxidation states of $$+2$$ and $$+4$$. This is because of extral stablility of fully-filled and half-filled orbitals.

Actinoids exhibit a greater range of oxidation states. This is because the $$5f, 6d$$ and $$7s$$ levels are of comparable energies. Again, $$+3$$ is the principal oxidation state for actinoids. Actinoids such as lanthanoids have more compounds in $$+3$$ state than in $$+4$$ state.

Explain giving reasons:
Transition metals and their many compounds act as good catalyst.

The catalytic activity of transition metals is ascribed to their ability to adopt multiple oxidation states and to form complexes.

Catalyst at a solid surface involves the formation of bonds between reactant molecules and atoms of the surface of the catalyst. This has the effect of increasing the concentration of the reactants at the catalyst surface and also weakening of the bonds in the reacting molecules (activation energy is lowered).

Transition metals have d and s orbitals to form these bonds.

Write balanced equations for the following :
NaCl is heated with sulphuric acid in the presence of $$ MnO_2 $$.

$$ 4 NaCl + MnO_2 + 4H_2SO_4 \rightarrow MnCl_2 + 4NaHSO_4 + 2H_2O + Cl_2 $$

Assign reasons for the following:
The transition metals and many of their compounds act as good catalyst.

Many transition metals and their compounds act as catalyst. The catalytic activity is due to their ability to exhibit multiple oxidation states. Example: $${V}_{2}{O}_{5}$$ in contact process and finely divided iron in Haber process.

When a magnetic field is applied, certain substances are attracted by the applied field. Those are called paramagnetic substances. Which is more paramagnetic $${Mn}^{2+}$$ or $${Cu}^{2+}$$? Why?

As the number of unpaired electrons increases, paramagnetism increases

$$Mn\rightarrow [Ar]4{s}^{2}3{d}^{5}$$ and $${Mn}^{2+}\rightarrow 3{d}^{5}$$

$$Cu\rightarrow [Ar]4{s}^{1}3{d}^{10}$$ and $${Cu}^{2+}\rightarrow 3{d}^{9}$$

$${Mn}^{2+}$$ contains five unpaired d-electrons and

$${Cu}^{2+}$$ contains only unpaired electron.

$$\therefore$$ $${Mn}^{2+}$$ is more paramagnetic than $${Cu}^{2+}.$$

Total number of molecules which hydrolysed at room temperature and hybridization of central atom in $$sp^{3}d$$ in transition state :
$$CCl_{4}, SiCl_{4}, NCl_{3}, PCl_{3}, AsCl_{3}, SF_{6}, P_{4}O_{6}, P_{4}O_{10}, SeF_{6}$$

Species are : $$SiCl_{4}, PCl_{3}, AsCl_{3}, P_{4}O_{6}, P_{4}O_{10}$$

So, number of such molecules is 5

$$(NH_4)_2Cr_2O_7$$ is the compound which imparts odour in fire works.
If true enter 1, else enter 0.

$$(NH_4)_2Cr_2O_7$$ may decompose to give $$NH_3$$ during fire works. Hence, it could give the odour of $$NH_3$$.
When ammonium dichromate is heated rapidly up to relatively low temperatures $$(498–503\ K)$$, the main product of an explosive reaction is $$Cr_2O_3$$ with some $$NO_2^-$$ and $$NH_3$$ trapped in the solid. With gradual heating of ammonium dichromate, even up to $$693\ K$$, the reaction is not a violent one. Instead, the colour of the reactant changes from orange to red, to brown and finally to black. $$Cr_2O_3$$ is formed once again with $$NO_2^-$$ and $$NH_3$$ as surface species. The existence of either a $$CrO_3$$, $$NH_3$$ or a $$Cr$$ and $$NH_3$$ complex is possible after gradual heating of $$(NH_4)_2Cr_2O_7$$ upto $$613\ K$$.

We use $$ZnO$$ in paints instead of $$Pb^{2+}$$ salts.
If true enter 1, else enter 0.

$$ZnO$$ in white paint by atmospheric $$H_2S$$ gives $$ZnS$$ (white) but $$Pb^{2+}$$ salts in white paint turns black by atmospheric $$H_2S$$ due to formation $$PbS$$. This is why we use ZnO instead of $${Pb}^{2+}$$ salts.
$$ZnO + H_2S \rightarrow ZnS (white) + H_2O$$
whereas, $$Pb^{2+} + H_2S \rightarrow PbS ( black) + 2H^+$$

Magnetic moment of $$Sc^{2+}$$ = Magnetic moment of $$Cu^{2+} $$
If true enter 1, else enter 0.

As $${Sc}^{2+}$$ and $${Cu}^{2+}$$ has same number of unpaired electrons, i.e. 1 unpaired electron due to having $$d^1$$ and $$d^9$$ configurations respectively, they also have same magnetic moment.
Hence , the magnetic moment $$= \sqrt {n(n+2)}) = \sqrt {1 ( 1 + 2 )} = \sqrt 3 = 1.73$$ BM.
So, both have same magnetic moment equal to 1.73 BM.

Match column I with column II and select the correct answer using the codes given below the columns.

Column I lists the metal ion and Column II lists the Magnetic moment (BM) :

The number of unpaired electron for $$ Cr^{3+}, Fe^{2+}, Ni^{2+} and \ Mn^{2+}$$ is 3, 4, 2 and 5 respectively.

Magnetic moment calculated by $$ \sqrt {n(n+2)}$$.

Magnetic moment of $$ Cr^{3+} $$= $$ \sqrt {3(3+2)}$$ = $$ \sqrt {15}$$.

Magnetic moment of $$ Fe^{2+} $$ = $$ \sqrt {4(4+2)}$$ = $$ \sqrt {24}$$.

Magnetic moment of $$ Ni^{2+} $$ = $$ \sqrt {2(2+2)}$$ = $$ \sqrt {8}$$.

Magnetic moment of $$ Mn^{2+}$$ = $$ \sqrt {5(5+2)}$$ = $$ \sqrt {35}$$.

Match the columns:

$${Ni}_{(28)}-{1s}^{2}{2s}^{2}{2p}^{6}{3s}^{2}{3p}^{6}{3d}^{8}{4s}^{2}$$
Number of unpaired electron in $${Ni}^{2+}=2$$
$$m=0$$ present in all orbitals
$$2+2+2+2+2+2+1=13(min)$$
Atom is paramagnetic.
It is transition element.

(b) $${K}_{(19)}-{1s}^{2}{2s}^{2}{2p}^{6}{3s}^{2}{3p}^{6}{4s}^{1}$$
$$m=0$$
Number of unpaired electron in $${K}^{2+}=1$$
$$m=0$$ present in all orbitals
$$2+2+2+2+2+1=11$$(min)
Atom is paramagnetic. It is not transition element.

(c) $${Mn}_{(25)}-{1s}^{2}{2s}^{2}{2p}^{6}{3s}^{2}{3p}^{6}{4s}^{2}{3d}^{5}$$
$${Mn}^{2+}$$
$$m=2+2+2+2+2+2+1=13$$
(Transition element)
Number of unpaired electron in $${Mn}^{2+}=5$$
$$m=0$$ present in all orbitals
$$2+2+2+2+2+2+1=13$$ (min)
Atom is paramagnetic. It is transition element.

(d) $$Pd-{1s}^{2}{2s}^{2}{2p}^{6}{3s}^{2}{3p}^{6}{4s}^{2}{3d}^{10}$$
$${Pd}^{2+} {4d}^{6}{5s}^{2}{4d}^{8}$$
Number of unpaired electron in $${Pd}^{2+}=2$$
$$m=0$$ present in all orbitals
$$m=2+2+2+2+2+2+2+2+2+1=19$$
Atom is diamangetic. It is transition element.

State whether the given statement is true or false:

Dipositive zinc exhibits paramagnetism due to loss of two electrons from $$3d$$-orbitals of neutral atom.

$$ Zn^{2+}$$ losses electrons from the $$ 4s^2$$ orbital and does not have any unpaired electron. Therefore it is diamagnetic.
$${zn}^{2+}$$

The oxidation number of $$Zn$$ is $$+2$$.

$$z=30$$

$$Zn=[Ar]{3d}^{10}{4s}^{2}$$

$${Zn}^{2+}=[Ar]{3d}^{10}$$

$$\uparrow \downarrow $$

No unpaired electron present; hence diamagnetic.

Hence,given statement is false.

Explain given reasons:

(i) Transition metals and many of their compounds show paramagnetic behaviour.
(ii) The enthalpies of atomisation of the transition metals are high.
(iii) The transition metals generally form coloured compounds.
(iv) Transition metals and their many compounds act as good catalyst.

(i) Transition metals and many of their compounds show paramagnetic behaviour.
This is due to presence of one or more unpaired electrons in d subshell.

(ii) The enthalpies of atomisation of the transition metals are high.
This is due to high effective nuclear charge and large number of valence electrons. This results in formation of strong metallic bonds.

(iii) The transition metals generally form coloured compounds.
This is due to d-d transition of unpaired electrons. In presence of ligands, the d orbitals split into two sets. Transition metal ions absorb radiation of a particular wavelength and reflect the remaining. This imparts colour.

(iv) Transition metals and their many compounds act as good catalyst.
Transition metals show variable oxidation states and forms complexes. They form unstable intermediate compounds. They provide a new path with lower activation energy of reaction. They also provide a suitable surface for the reaction to occur.

The chemistry of the actinoids is not so smooth as that of the lanthanoids. Justify this statement by giving some examples from the oxidation state of these elements.

Apart from +3 oxidation state, lanthanoids show +2 and +4 oxidation states due to large energy gap between 4f and 5d subshells. Whereas actinoids show large number of oxidation states due to small energy gap between 5f and 6d subshells.
For example, uranium (Z = 92) shows +3, +4, +5 and +6 oxidation states and neptunium (Z=94) shows +3,+4,+5,+6 and +7 oxidation states due to small energy gaps between 5f and 6d orbitals.

Use Hund's rule to derive the electronic configuration of $$Ce^{3+}$$ ion and calculate its magnetic moment on the basis of spin-only formula.

$$\displaystyle Ce (Z = 58) \: : \: [Xe]^{54} 4f^15d^16s^2$$

$$\displaystyle Ce^{3+} \: : \: [Xe]^{54} 4f^1 $$

It has $$1$$ unpaired electron.

Magnetic moment, $$\displaystyle \mu _s = \sqrt {n(n+2)} = \sqrt {1(1+2)} = \sqrt {3} = 1.73 $$ B.M

Compare the chemistry of actinoids with that of the lanthanoids with special reference to: (i) electronic configuration (ii) atomic and ionic sizes and (iii) oxidation state (iv) chemical reactivity.

(i) Electronic configuration
lanthanoids have general electronic configuration $$\displaystyle [Xe]4f^{1-14}5d^{0-1}6s^2$$. Actinoids have general electronic configuration $$\displaystyle [Rn]5f^{1-14}6d^{0-1}7s^2 $$.
In lanthanoids, 4f orbital is filled and in actinoids, 5f orbital is filled.
(ii) Atomic and ionic sizes
The atomic and ionic sizes of lanthanoids and actinoids decrease with increase in the atomic number due to lanthanoid and actinoid contraction.
(iii) Oxidation state
Apart from +3 oxidation state, lanthanoids show +2 and +4 oxidation states due to large energy gap between 4f and 5d subshells. Whereas actinoids show large number of oxidation states due to small energy gap between 5f and 6d subshells.
(iv) Chemical reactivity
Highly electropositive lanthanoids have almost similar chemical reactivity. Actinoids (electropostive and highly reactive) are more reactive (specially in finely divided state) than lanthanoids.

What are alloys? Name an important alloy which contains some of the lanthanoid metals. Mention its uses.

An alloy is a solid or solution of two or more elements in a metallic matrix. It can either be a partial solid solution or a complete solid solution. Alloys are usually found to posses different physical properties than those of the component elements.

An important alloy of lanthanoids is Mischmetal. It contains lanthanoids (4-95%), iron (5%) and traces of $$S,C,Si, Ca$$ and $$Al$$

Uses

(1) Mischmetal is used in cigarettes and gas lighters

(2) It is used in flame throwing tanks

(3) It is used in flame bullets and shills.

Predict which of the following will be coloured in aqueous solution? $$Ti^{3+}, V^{3+}, Cu^{+}, Sc^{3+}, Mn^{2+}, Fe^{3+}$$ and $$Co^{2+}$$. Give reasons for each

Transition metal ions with incompletely filled d orbitals are colored and those with completely filled d orbitals (or vacant d orbitals) are colorless.
$$\displaystyle Ti^{3+}, V^{3+}, Mn^{2+}$$ and $$\displaystyle Co^{2+}$$ are colored due to d-d transition in incompletely filled d orbitals.
$$\displaystyle Cu^{+}, Sc^{3+} $$ are colorless due to completely filled and empty d orbitals respectively.

Compare the general characteristics of the first series of the transition metals with those of the second and third series metals in the respective vertical columns. Give special emphasis on the following points:
(i) electronic configurations (ii) oxidation states (iii) ionisation enthalpies and (iv) atomic sizes

(i) Electronic configuration
In the first transition series, 3d orbitals are progressively filled whereas, in second and third transition series, 4d and 5d orbitals are filled.

(ii) Oxidation states
+2 and +3 oxidation states of first transition series elements are more stable than +2 and +3 oxidation states of second and third transition series elements. Cobalt forms several complexes with +2 and +3 oxidation states but no such complexes are known for $$Rh, Ir$$ which belong to the same group to which $$Co$$ belongs. The maximum oxidation state in a group increases from first transition series to third transition series. In group 8, iron shows +2 and +3 oxidation states and ruthenium and osmium show +4, +6 and +8 oxidation states.

(iii) Ionization enthalpies
5d series elements have higher ionization enthalpies than 3d and 4d series elements. In 5d series, 4f orbitals are filled which have poor shielding effect and valence electrons experience higher effective nuclear charge. This results in higher ionization enthalpies for 5d series elements.

(iv) Atomic sizes
4d series elements have higher atomic size than 3d series elements.
4d and 5d series elements have nearly same atomic size due to lanthanoid contraction.

Compare the chemistry of the actinoids with that of lanthanoids with reference to: (i) electronic configuration (ii) oxidation states and (iii) chemical reactivity

(i) Electronic configuration
Lanthanoids have general electronic configuration $$\displaystyle [Xe]4f^{1-14}5d^{0-1}6s^2$$. Actinoids have general electronic configuration $$\displaystyle [Rn]5f^{1-14}6d^{0-1}7s^2 $$ .
In lanthanoids, 4f orbital is filled and in actiniods, 5f orbital is filled.
(ii) Oxidation state
Apart from +3 oxidation state, lanthanoids show +2 and +4 oxidation states due to large energy gap between 4f and 5d subshells. Whereas actinoids show large number of oxidation states due to small energy gap between 5f and 6d subshells.
(iii) Chemical reactivity
Highly electropositive lanthanoids have almost similar chemical reactivity. Actinoids (electropostive and highly reactive) are more reactive (specially in finely divided state) than lanthanoids.

Comment on the statement that elements of the first transition series possess many properties different from those of heavier transition elements.

The properties of the elements of the first transition series differ from those of the heavier transition elements in many ways:

(i) The atomic sizes of the elements of the first transition series are smaller than those of the heavier elements of 2nd and 3rd transition series. However, the atom sizes of the elements in the third transition series are virtually the same as those of the corresponding members in the second transition series this is due to lanthanoid configuration.

(ii) +2 and +3 oxidation states are more common for elements in the first transition series, while higher oxidation states are more common for the heavier elements

(iii) The enthalpies of atomisation of the elements in the first transition series are lower than those of the corresponding elements in the second and third transition series

(iv) The melting and boiling points of the first transition series are lower than those of the heavier transition elements. This is because of the occurrence of strong metallic bonding (M-M bonding)

(v) The elements of the first transition series from low-spin or high-spin complexes depending upon the strength of the ligan field. However, the heavier transition elements form only low-spin complexes, irrespective of the strength of the ligand field.

Write one similarity and one difference between the chemistry of lanthanoid and actinoid elements

Similarity : In lanthanoids and actinoids filling of (n-2)f orbitals is involved.

Difference : Lanthanides can show a maximum oxidation state of +4 while actinides show oxidation states of +3, +4, +5, +6 and +7.

Fourteen elements following Lanthanum are called Lanthanoids:
(a) What is Lanthanoid contraction? Give reason for it.
(b) $$KMn{O}_{4}$$ is a purple coloured crystal and it acts as an oxidant. How will you prepare $$KMn{O}_{4}$$ from $$Mn{O}_{2}$$?

Lanthanide Contraction: The steady decrease in the size of lanthanide ions (Tri positive metal ion) with the increase in atomic no. is called lanthanide contraction.

Reason: As we move from left to right in a lanthanide series, the atomic number increases and for every proton in the nucleus the new electron goes to 4f orbital. The 4f orbital is too diffused to protect the nucleus effectively, thus there is a gradual increase in the effective nuclear charge experienced by the outer electrons. Consequently, the attraction of the nucleus for the electrons in the outermost shell increases with the increase of atomic number, so the size decreases.

b) Potassium permanganate ($$KMnO_4$$) is prepared from pyrolusite ore ($$MnO_2$$). The finely powdered $$MnO_2$$ is fused with an alkali metal hydroxide like KOH in the presence of air to give the potassium Manganate (K2MnO4)

$$MnO_2 + 4 KOH + O_2 \rightarrow 2K_2MnO_4 + 2H_2O$$

Account for the following:
(i) Transition metals show variable oxidation states.
(ii) $$Zn, Cd$$ and $$Hg$$ are soft metals.
(iii) $$E^{\circ}$$ value of the $$Mn^{3+}/Mn^{2+}$$ couple is highly positive $$(+ 1.57\ V)$$ as compared to $$Cr^{3+}/Cr^{2+}$$

(a) Transition elements show variable oxidation state due to very small energy difference in (n-1)d and ns-orbitals

(b) The electronic configurations of Zn, Cd and Hg are represented by the general formula (n-1)d¹⁰ns². The orbitals in these elements are completely filled in the ground state as well as in their common oxidation states

(c) $$Mn^{2+}$$ exists in $$3d^5$$ configuration which is half filled configuration and such half filled configurations provide extra stability to the Metal Ion. While $$Mn^{3+}$$ will exist in $$3d^4$$ configuration which is less stable than $$3d^5$$. Because of $$3d^3$$ half filled d orbital extra stability is attained by $$Cr^{+3}$$ than $$Cr^{+2}$$. hence $$Cr^{+3}\rightarrow Cr^{+2}$$ is less feasible.

Explain why:
$$Cu^{+}$$ is diamagnetic but $$Cu^{2+}$$ is paramagnetic. $$(Z = 29)$$

$$Cu^{+}$$ is diamagnetic because it has no unpaired electrons in its core.
On the other hand $$Cu^{++}$$ is paramagnetic because it has an unpaired electron in its core.

Why do transition elements form complexes?

Explanation:
Part: 1

The d-block elements are generally called transition elements.

The general electronic configuration is $$ns^{2}\left ( n-1 \right )d^{1-10}$$.

Part: 2 Reasons:

Transition metals more often form complexes by bonding with neutral or anionic molecules or elements because:

Small size
High nuclear charge
Availability of empty d-orbitals to accommodate the lone pairs of electrons of the ligands.

Which of the following arrangments does not represent the correct order of the property stated against it?
(1) $$V^{2+} < Cr^{2+} < Mn^2 < Fe^{2+}$$ : paramagnetic behaviour
(2) $$Ni^{2+} < Co^{2+} < Fe^{2+} < Mn^{2+}$$ : ionic size
(3) $$Co^{3+} < Fe^{3+} < Cr^{3+} < Sc^{3+}$$ : stability in aqueous solution
(4) $$Sc < Ti < Cr < Mn$$ : number of oxidation states

a) Paramagnetic behavoiur :

$${ V }^{ 2+ }\left( { 3d }^{ 3 } \right) \rightarrow n=3$$

$${ Cr }^{ 2+ }\left( { 3d }^{ 4 } \right) \rightarrow n=4$$

$${ Mn }^{ 2+ }\left( { 3d }^{ 5 } \right) \rightarrow n=5$$

$${ Fe }^{ 2+ }\left( { 3d }^{ 0 } \right) \rightarrow n=4$$

$$\Rightarrow { Mn }^{ 2+ }>{ Fe }^{ 2+ }>{ Cr }^{ 2+ }>{ V }^{ 2+ }$$

b) $${ Ni }^{ 2+ }<{ CO }^{ 2+ }<{ Fe }^{ 2+ }<{ Mn }^{ 2+ }$$ : ionic size

For same O.S, the ionic radii generally decreases with increase in nuclear charge and nuclear charge increases as we move across period.

c) $${ CO }^{ 3+ }<{ Fe }^{ 3+ }<{ Cr }^{ 3+ }<{ SC }^{ 3+ }$$ : stability in aq. solutions the lowest the electrode potential, the more stable is the transition metal in aq. solution. The value becomes less $$-ve$$ across the series.

d) $$Sc<Ti<Cr<Mn$$ : number of oxidation states.

Consider the following list of reagents:
Acidified $${K}_{2}{Cr}_{2}{O}_{7}$$, alkaline $$KMn{O}_{4},Cu{SO}_{4},{H}_{2}{O}_{2},{Cl}_{2},{O}_{2},Fe{Cl}_{2},H{NO}_{3}$$ and $${Na}_{2}{S}_{2}{O}_{3}$$. The total number of reagents that can oxidise aqueous iodide to iodine is:

Answer is 5

$$K_2Cr_2O_7 +I^- +H^+\rightarrow I_2$$

$$CuSO_4+I^- \rightarrow I_2$$

$$H_2O_2 +I^- \rightarrow I_2$$

$$Cl_2+I^- \rightarrow I_2$$

$$HNO_3 +I^- \rightarrow I_2$$

Match the compounds/elements given in Column 1 with uses given in Column 2.
Column 1:- Column 2:-
I) Lanthanoid oxide. A) production of iron alloy
II) Lanthanoid. B) television screen
III) Misch metal. C) petroleum cracking
IV) Magnesium based alloy D) Lanthanoid metal+iron
is constituent of
V) Mixed oxides of Lanthanoids E) bullets
are employed
F) in X-ray screen

2105930_1178020_ans_d5c8ed3698f44f4e989648c588facaf4.jpg

Give reason :
$$[Ti(H_{2}O)_{6}]^{3+}$$ is coloured while $$[Sc(H_{2}O)_{6}]^{3+}$$ is colourless

$$[Ti(H_{2}O)_{6}]^{3+}$$ is coloured while $$[Sc(H_{2}O)_{6}]^{3+}$$ is colourless This is because, in $$Ti(H2O)_6$$ has unpaired electron for d-d transition( responsible for colorness of a complex compound) so it is coloured .while $$Sc(H2O)_6$$ can not undergo d-d Transition as do not have unpaired electrons so it is colorless.

Why salts of $$Sc^{3+},Ti^{4+},V^{5+}$$ are colourless?

1332950_1317954_ans_390f86066f9a4556a5e401be5b9efe94.jpg

Indicate the steps in the preparation of $$ \mathrm { K } _ { 2 } \mathrm { Cr } _ { 2 } \mathrm { O } _ { 7 } $$ from chromite ore.

Potassium dichromate: ($${K}_{2}{Cr}_{2}{O}_{7}$$) is prepared from chromite ore $$(Fe{Cl}_{2}{O}_{4})$$ in the following steps:

(1) Preparation of sodium chromate

$$4Fe{Cr}_{2}{O}_{4}+16NaOH+7{O}_{2}\rightarrow 8{Na}_{2}Cr{O}_{4}+2{Fe}_{2}{O}_{3}+8{H}_{2}O$$

(2) Conversion of sodium chromate into sodium dichromate

$$2{Na}_{2}Cr{O}_{4}+conc.{H}_{2}{SO}_{4}\rightarrow {Na}_{2}{Cr}_{2}{O}_{7}+{Na}_{2}{SO}_{4}+{H}_{2}O$$

(3) Conversion of sodium dichromate to potassium dichromate

$${Na}_{2}{Cr}_{2}{O}_{7}+2KCl\rightarrow 2NaCl+{K}_{2}{Cr}_{2}{O}_{7}$$

Potassium chloride being less soluble than sodium chloride is obtained in the form of orange coloured crystals and can be removed by filteration.

The dichromate ion ($${ Cr }_{ 2 }{ O }_{ 7 }^{ 2- }$$) exists in equilibrium with chromate $$(Cr{O}_{4}^{2-})$$ ion at pH.$$4$$. However, by changing the pH they can be interconverted.

$$\underset { chromate\\ (yellow) }{ 2Cr{ O }_{ 4 }^{2-} } \quad \overset { acid }{ \underset { alkali }{ \leftrightharpoons } }\quad \underset { Hydrogen\\ chromate }{ 2HCr{ O }_{ 4 } ^-} \quad \overset { acid }{ \underset { alkali }{ \leftrightharpoons } }\quad \underset { Dichromate\\ (orange) }{ { Cr }_{ 2 }{ O }_{ 7 }^{ 2- } } $$

What is lanthanoid contraction? What are the consequences of lanthanoid contraction?

The regular decrease in atomic radii due to the filling of $$4f$$ before $$5d$$ orbitals is called lanthanoid contraction. This is because of the imperfect shielding of one electron by another in the same set of orbitals. The shielding of one $$4f$$ electron by another is less than that of one $$d$$ electron by another, and as the nuclear charge increases along with the series, there is a fairly regular decrease in the size of the entire $$4f$$ orbitals.

Consequences of lanthanoid contraction:

i. Radii of the third series are virtually the same as those of the corresponding member of the second series.

ii. Third series and second series have very similar physical and chemical properties much more than that expected by usual family relationships.

Complete the following reactions:

? $$ \underset {[H^{+}]}{\xrightarrow {KMnO_{4}}}CH_{3}CH_{2}\overset {\overset {O}{\parallel}}{C}CH_{3} + HO - \overset {\overset {O}{\parallel}}{C} CH_{2} CH_{2}CH_{2}CH_{3}$$.

$$CH_{3} - CH_{2} - \underset {\underset {CH_{3}}{|}}{C} = CH - CH_{2} - CH_{2} - CH_{2} - CH_{3}$$.

Describe the structure of interstitial and substitutional alloys and outline the factor determining which is formed.

Structure of some reactants and products are given in column I and II respectively. First match the appropriate reactant-product pair from column I and II and then suggest another suitable reagent for each pair:

1546121_1722537_ans_d681272218df4e99ac615afdff6a0358.jpg

When orange solution containing $$ Cr_2O_7^{2-} $$ ion is treated with an alkali, a yellow solution is formed and when $$ H^+ $$ ions are added to yellow solution,an orange solution is obtained. Explain why does this happen?

2032240_1724096_ans_5d5192d9be2747c6b50aaa58ba559033.jpg

When a brown compound of manganese (A) is treated with $$HCl$$ it gives a has (B). The gas taken is excess, reacts with $$ NH_3 $$ to give an explosive compound (C). Identify compounds A, B, and C.

2105777_1724085_ans_36ebab55d7b847dda61f1ec2aea8b542.jpg

Find the magnetic moment of a salt containing $$Zn^{2+}$$ ion.

1881374_1725341_ans_0a3d96eb976844d6b91f2f9d6dbee54a.jpeg

An aqueous blue-colored solution of a transition metal suphate reacts with $$ H_2S $$ in acidic medium to give a black precipitate A which is insoluble in warm aqueous solution of KOH. The blue solution on treatement with KI in weakly acidic medium turns yellow and produces a white precipitate B. Identify the transition metal ion. Write the chemical reactions involved in the formation of A and B.

1577795_1724728_ans_3bef580a20ec4ed2b83b3cffe465cc1f.jpg

Complete the following equation:
$$Cr_{2}O_{7}^{2-}+H^{+}+SO_{2}\rightarrow SO_{4}^{2-}+H_{2}O+......$$

The balanced chemical equation is :

$$Cr_{2}O_{7}^{2-}+2H^{+}+3SO_{2}\rightarrow 3SO_{4}^{2-}+H_{2}O+2Cr^{3+}$$

Among $$PbS,CuS,HgS,MnS,Ag_2S,NiS,CoS,Bi_2S_3$$ and $$SnS_2$$ the total number of BLACK coloured sulphides is:

$$PbS,CuS,HgS,Ag_2S,NiS,CoS$$ are black

$$MnS-$$ dirty pink/Buff

$$SnS_2-$$ yellow

$$Bi_2S_3-$$ brown/black (brownish black)

Explain the following:
$$[CoF_6]^{3-}$$ is paramagnetic but $$[Co(CN)_6]^{3-}$$ is diamagnetic.

1962007_1729828_ans_96ed8f063657473fa0b586b0d2e3cea7.jpg

$$[FeCN)_6]^{3-}$$ ion has magnetic moment of $$1.41\ B.M$$. While $$[Fe(H_2O)_6]^{3+}$$ has a magnetic moment of $$5.92\ B.M.$$ Explain.

Magnetic moment, $$\mu=[n(n+2)]^{1/2}$$ (n=number of unpaired electrons)

In $$[Fe(CN)_6]^{-3}$$

Oxidation state of $$Fe$$ = $$+3$$

$$CN$$ being a strong field ligand, pairs up electrons in $$3d$$ orbital of $$Fe$$. It is a low spin complex.

Therefore, it is $$d^2sp^3$$ hybridized (Octahedral structure)

Total number of unpaired electrons, $$n = 1$$

$$\mu=[1(1+2)]^{1/2}$$ = $$1.41\space BM$$

In $$[Fe(H_2O)_6]^{+3}$$

Oxidation state of $$Fe$$ = $$+3$$

$$H_2O$$ being a weak field ligand, do not pair up electrons in $$3d$$ orbital of $$Fe$$. It is a high spin complex.

Therefore, it is $$sp^3d^2$$ hybridized (Octahedral structure)

Total number of unpaired electrons, $$n = 5$$

$$\mu=[5(5+2)]^{1/2}$$ = $$5.92\space BM$$

Hence $$[Fe(CN)_6]^{-3}$$ ion has magnetic moment 1.41 BM while $$[Fe(H_2O)_6]^{+3}$$ has a magnetic moment of 5.92 BM.

1581761_1731660_ans_cfa06cd0a0574b8bbf894e2497796d68.jpg

Match the List $$I$$ with their characteristic List in $$II$$:

1728784_1731827_ans_434e87a9d2c14ef790c55ac35a6b342d.jpg

On the basis of lanthanoid contraction, explain the nature of bonding in $$La_{2}O_{3}$$ and $$Lu_{2}O_{3}$$.

1659709_1778873_ans_7ac9f4f4434d4d369c9c7d53e7c48873.jpg

Why do transition elements form complexes ?

Transition metals generally form more complex or coordination compound because they have empty valence shell orbitals that can accept pair of an electron from lewis base (ligand) That means ligands must contain one pair (at least) of the non-bonding electron that can be donated to the metal ion.

Match the compounds in Column I with its characteristics reactions given in Column II.

1788308_1769991_ans_5b85941a5bac4efda228961e0bdb922c.png

Some d-block elements are given below:
$$(Cr,Mn,Fe,Co,Ni)$$
(a) Identify the element which shows maximum paramagnetic behaviour
(b) Give reason for the highest paramagnetic property of that element
(c) When two electrons are lost from each atom, is there any change in magnetic properties of each? Explain.

(a) $$Cr$$ is the element which shows maximum paramagnetic behaviour.

(b)

$$Cr-4{s}^{1}3{d}^{5}\to $$ 6 unpaired electrons

$$Mn-4{s}^{2}3{d}^{5}\to$$ 5 unpaired electrons

$$Fe-4{s}^{2}3{d}^{6}\to $$ 4 unpaired electrons

$$Co-4{s}^{2}3{d}^{7}\to $$ 3 unpaired electrons

$$Ni-4{s}^{2}3{d}^{8}\to $$ 2 unpaired electrons

Since there are six unpaired electrons in $$Cr$$, it has the maximum paramagnetic behaviour

(c)

$${Cr}^{2+}\rightarrow 3{d}^{4}\to$$ 4 unpaired electrons

$${Mn}^{2+}\rightarrow 3{d}^{5}\to$$ 5 unpaired electrons

$${Fe}^{2+}\rightarrow 3{d}^{6}\to$$ 4 unpaired electrons

$${CO}^{2+}\rightarrow 3{d}^{7}\to$$ 3 unpaired electrons

$${Ni}^{2+}\rightarrow 3{d}^{8}\to$$ 2 unpaired electrons

When two electrons are lost, $${Mn}^{2+}$$ contains the maximum number of unpaired electrons i.e. $${Mn}^{2+}$$ is the ion with maximum paramagnetic behaviour. $${Cr}^{2+}$$ and $${Fe}^{2+}$$ show similar paramagnetism and so on.

Write the uses of lanthanides and actinides.

Both lanthanides and actinides are radio active so mainly they are used to generate power in nuclear power plant through the process of fission or breaking apart. This power is used to generate electricity , in medical application such as x-ray and to create bombs.

Also they are used to create alloy or a metal that is made by mixing more than one metal element to increase magnetism or strength of material.

Write electronic configuration of $${Cu}^{2+}$$. Calculate its magnetic moment.

Electronic configuration of $${Cu}^{2+}$$ will be $$[Ar]3{d}^{9}$$

so magnetic moment is given by

$$\mu =\sqrt {n(n+2)}$$

where $$n=1$$

(No. of unpaired electrons)

$$\mu =\sqrt{1(1+2)}=\sqrt{3}=1.73BM$$

Describe the preparation of potassium dichromate from iron chromite ore. What is the effect of increasing pH on a solution of potassium dichromate?

Potassium Dichromates are generally prepared from chromates which in turn are obtained by the fusion of chromite ore $$(Fe{Cr}_{2}{O}_{4})$$ with sodium or potassium carbonate in free excess of air.

$$4Fe{Cr}_{2}{O}_{4}+8{Na}_{2}{CO}_{3}+7{O}_{2}\rightarrow 8{Na}_{2}Cr{O}_{4}+2{Fe}_{2}{O}_{3}+8{CO}_{2}$$

The yellow solution of $${Na}_{2}Cr{O}_{4}$$ is filtered and acidified with $${H}_{2}{SO}_{4}$$ to give a solution from which orange sodium dichromate $${Na}_{2}{Cr}_{2}{O}_{7}.2{H}_{2}O$$ can be crystallised.

$$2{Na}_{2}Cr{O}_{4}+2{H}^{+}\rightarrow {Na}_{2}{Cr}_{2}{O}_{7}+2{Na}^{+}+{H}_{2}O$$

Sodium dichromate is more soluble than potassium dichromate. The latter is prepared by treating the solution of $${Na}_{2}{Cr}_{2}{O}_{4}$$ with $$KCl$$.

$$Na_{2}Cr_{2}O_{7}+2KCl\rightarrow {K}_{2}{Cr}_{2}{O}_{7}+2NaCl$$ (orange crystals)

On increasing the pH of a solution of $${K}_{2}{Cr}_{2}{O}_{7}$$, it behaves like a strong oxidising agent and it oxidises iodide to iodine sulphides to sulphur, tin (II) to tin (IV) and iron (II) salts to iron (III) salts.

The magnetic moment of a compound of transition element is $$3.9B.M$$. Write the number of unpaired electrons in the element.

Magnetic moment of any compound is given by

$$\mu=\sqrt{n(n+2)}$$

where $$n$$ is the no. of unpaired electrons

So,

$$3.9=\sqrt{n(n+2)}$$

squaring both sides

$$15.21={n}^{2}+2n....(1)$$

This equation (1) satisfying for $$n=3$$. So, no. of unpaired electrons in this elements will be three.

Write four differences between lanthanoids and actinoids.

Property	Lanthanoids	Actinoids
1. Binding energies	Binding energies of 4f orbitals are higher.	Binding energies of 5f orbitals are lower.
2. Radioactivity	Except, promethium, these are non-radioactive	All actinoids are radioactive
3. Basic character	Lanthanoid compounds are less basic.	Actinoid compounds are more basic.
4. Tendency to form oxo ions	They do not form oxo-ions.	They form oxo ions such as $${UO}_{2}^{2+},Np{O}_{2}^{2+},Pu{O}_{2}^2{+}$$

Predict which of the following will be coloured in aqueous solution?

$${Ti}^{3+},{V}^{3+},{Cu}^{+},{Sc}^{3+},{Mn}^{2-},{Fe}^{3+}$$ and $${Co}^{2+}$$

Give reasons for each:

Metal ions	Electronic configuration
$${Ti}^{3+}$$	$$[Ar]3{d}^{1}4{s}^{0}$$
$${V}^{3+}$$	$$[Ar]3{d}^{2}4{s}^{0}$$
$${Cu}^{+}$$	$$[Ar]3{d}^{10}4{s}^{0}$$
$${Sc}^{3+}$$	$$[Ar]3{d}^{0}4{s}^{0}$$
$${Mn}^{2+}$$	$$[Ar]3{d}^{5}4{s}^{0}$$
$${Fe}^{3+}$$	$$[Ar]3{d}^{5}4{s}^{0}$$
$${Co}^{2+}$$	$$[Ar]3{d}^{7}4{s}^{0}$$

Among the above mentioned ions $${Ti}^{3+},{V}^{3+},{Cu}^{+},{Sc}^{3+},{Mn}^{2+},{Fe}^{3+}$$ and $${Co}^{2+}$$ will be coloured in its aqueous solution due to the ability of electron to jump from a lower energy $$d$$ orbital to a higher energy $$d-$$ orbital.

In case of the ions $${Cu}^{+},{Sc}^{3+}$$, this d-d transition cannot take place either due to the absence of any electrons in 3d orbital or due to complete filling of d orbital

Assign A, B, C, D from given type of reactions.
A for precipitate formation reaction.
B for precipitate dissolution reaction.
C for precipitate exchange reaction.
D for no reaction.
$$2BaCrO_4+4HCl \longrightarrow 2BaCl_2+H_2Cr_2O_7+H_2O$$

It is a precipitate dissolution reaction, with the ppt as $$BaCrO_4$$

The D-And F-Block Elements - Class 12 Engineering Chemistry - Extra Questions

The sum of the total number of bonds between chromium and oxygen atoms in chromate and dichromate ions is ________.

Explain why:Transition elements form coloured compounds.

................... gives blue colour to the Glass.

Why do transition elements form complexes?

Differentiate Lanthanides and Actinides.

For the dichromate dianion, which of the following is true?1) $$4 Cr - O$$ bonds are equivalent 2) $$6 Cr - O$$ bonds are equivalent

Transition metals are good catalytic agent. Why?

Represent the structure of dichromate ion.

What is the magnetic property of $$Ni$$ atom $$[Z=28]$$ ?

Name the method to determine magnetic properties.

How do you account for the larger size of lanthanum than hafnium though they are successive elements?

Magnetic moment of $$V^{2+}$$ = Magnetic moment of $$Co^{2+} $$ if true enter 1, if false enter 0.

The value of 3.00 BM into SI unit is $$x \times 3.09 \times 10 ^{-24} Am^2$$. The value of $$x $$ is ______.

The colour of $$MnO^-_4$$ ion is due to charge transfer.If true enter 1, else enter 0.

Give a structure of $$Cr_2O^{2-}_7$$ ion.

Angle $$\alpha $$ in degrees is:

Magnetic moment of $$Fe^{3+}$$ ions $$\sqrt {35}$$ BM. Determine number of unpaired electrons.

Titanium is much harder than potassium or calcium and so, it is used in making jet aircraft engines.If true enter 1, else enter 0.

Which of the following is not physically change ?Magnetizing of a piece of iron.

No. of peroxide bonds in blue perchromate is two.If true enter 1, else enter 0.

What do you observe when water droplets fall on blue cobalt chloride ?

The transition metals (with the exception of Zn, Cd and Hg) are hard and have high melting and boiling points.If true enter 1 else 0

Transition elements show variable oxidation states . If true enter 1 else 0.

What is meant by disproportionation? Give two examples of disproportionation reaction in aqueous solution.

Give one point of dissimilarity between lanthanides and actinides.

(a) Which metal in the first transition series (3d series) exhibits $$+ 1$$ oxidation state most frequently and why?(b) Which of the following cations are coloured in aqueous solutions and why?$$Sc^{3+}, V^{3+}, Ti^{4+},Mn^{2+}$$(At. Nos. $$Sc = 21,\ V = 23,\ Ti = 22,\ Mn = 25$$)

(a) How would you account for the following: (i) The chemistry of actinoids is more complicated as compared to lanthanoids (ii) Transition metals form complex compounds(b) Complete the following equation: $$2{Mn{ O }_{ 4 }}^{ - }+6{ H }^{ + }+S{{ O }_{ 3 }}^{ 2- }\longrightarrow $$

What type of magnetism is shown by a substance if magnetic moments of domains are arranged in same direction?

What are the transition elements ? Write two characteristics of the transition elements.

$$Ti^{4+}$$ ion is colourless. Give reason.

$$CuI_2$$: Brown :: $$AgCl$$ : ________

Give the equations for the conversion of argentite $$(Ag_2S)$$ to metallic silver.

Why is $$S{ c }^{ 3+ }$$ colourless while $${ Ti }^{ 3+ }$$ coloured ? (Atomic number $$Sc=21, Ti=22$$)

Explain the following:$$(i)$$ Why do transition metal ions possess a great tendency to form complexes?$$(ii)$$ The paramagnetic character in $$3d$$-transition series elements increases upto $$Mn$$ and then decreases.

What is "Spitting of Silver"? How is it prevented?

Write a note of Plumbosolvency.

How does gold react with aquaregia ?

Explain chromyl chloride test with equation.

Why do d-block elements from complexes?

Aqueous $$Cu^{+2}$$ ions are blue in color, where as aqueous $$Zn^{+2}$$ ions are colorless. Why?

What is tailing of Mercury?

Give the composition of the following alloys:(a) German silver(b) Brass

Write the characteristic properties of transition elements.

What is spitting of silver? How can it be prevented?

Give the composition and uses of Nichrome.

What is "Tailing of Mercury"

Write the uses of Lanthanides.

Write the equations involved in the preparation of potassium dichromate from chromite ore.

Explain:(i) Transition elements mostly form the complex compound. Why?(ii) Transition elements are good catalyst. Why?(iii) Transition elements show variable valency. Why?

Write the factors which are related to the colour of transition metal ions.

Explain the following (i) $$Cu^+$$ is colourless and $$cu^{2+}$$ is coloured,(ii) $$Zn$$ shows only + 2 Oxidation state.

With reference to the first row transition series :i) Name the metal which possesses maximum number of oxidation states.ii) Among $$Zn^{+2}$$ and $$Cu^{+2}$$ which is colourless?iii) Between $$Ti^2+$$ and $$V^{2+}$$ which ion contains more number of unpaired electrons?

(a) What are Lanthanide elements?(b) What do you understand by Lanthanide contraction?(c) Write chromyl chloride test with equation.

Give any two differences between lanthanoids and actinoids.

(a) Why Transition elements become paramagnetic?(b) Explain oxidization properties of potassium permanganate in acidic medium. (any four only)

Calculate the value of magnetic moment of $$V^{+2}$$ ?

Calculate the 'spin only' magnetic moment of $$M^{2+}(aq)$$ ion (Z=27).

What is lanthanide contraction? Explain the cause of lanthanide contraction.

(i) Scandium (z = 21) is a transition element but zink (z = 30) is not. Explain.(ii) Calculate the equivalent weight of $$KMnO_4$$in acidic medium.(iii) What do you mean by Lanthanoid contraction ?

Out of $$C{ r }^{ 3+ }$$ and $$M{ n }^{ 3+ }$$, which is stronger oxidising agent and why?

Why do transition elements form alloys?

Explain why:Zr and Hf exhibit similar properties.(At. no. of Zr = 40, Hf = 72)

What happens when sulphur dioxide gas is passed through an aqueous solution of a Fe(III) salt?

(i) Write down any three difference between Lanthanoid and Actinoid ?(ii) The melting and boiling points of Zy. Cd and Hg are low. Why ? (iii) Draw the structure of manganate ion.

Explain why:Transition elements usually form coloured ions.

What is meant by ferrimagnetism? Give any two examples.

Write any four applications of $$'f'$$ - block elements.

In the $$3$$d series from Sc$$(z=21)$$ to Zn$$(z=30)$$, enthalpy of atomisation of Zn is the lowest.

What happens when $$(NH_4)_2Cr_2O_7$$ is heated?

Magnetic momentum of an ion $${ M }^{ +3 }$$ is $$\sqrt { 35 }\ BM$$ Then, number of electron in $$d$$ orbital of $$m$$ element will be__________.

Which of the following ions is coloured in aqueous solution?i) $$Cu^{2+}$$ii) $$Cu^{1+}$$

Write the nomenclature and symbol of an element having atomic number 119?

What is magnetic moment of $${ V }^{ +4 }$$?

Why are metals like chromium and molybdenum very hard?

Name the transition element of highest density.

Why basic character in lanthanoid hydroxides decreases across the period?

Transition metals and their compounds are used as catalysts. Give two reasons.

Write a chemical equation for burning of lanthanoids ($$ln$$) in oxygen.

Use hund's rule to derive the electronic configuration of $$Ce^{+3}$$ ion, and calculate its magnetic moment on the basis of 'spin only' formula. ( For Ce, Z = 58)

Explain why:
Transition elements form coloured compounds.

For the dichromate dianion, which of the following is true?
1) $$4 Cr - O$$ bonds are equivalent
2) $$6 Cr - O$$ bonds are equivalent

The colour of $$MnO^-_4$$ ion is due to charge transfer.
If true enter 1, else enter 0.

Titanium is much harder than potassium or calcium and so, it is used in making jet aircraft engines.
If true enter 1, else enter 0.

Which of the following is not physically change ?
Magnetizing of a piece of iron.

No. of peroxide bonds in blue perchromate is two.
If true enter 1, else enter 0.

The transition metals (with the exception of Zn, Cd and Hg) are hard and have high melting and boiling points.
If true enter 1 else 0

(a) Which metal in the first transition series (3d series) exhibits $$+ 1$$ oxidation state most frequently and why?
(b) Which of the following cations are coloured in aqueous solutions and why?
$$Sc^{3+}, V^{3+}, Ti^{4+},Mn^{2+}$$
(At. Nos. $$Sc = 21,\ V = 23,\ Ti = 22,\ Mn = 25$$)

(a) How would you account for the following:
(i) The chemistry of actinoids is more complicated as compared to lanthanoids
(ii) Transition metals form complex compounds
(b) Complete the following equation:
$$2{Mn{ O }_{ 4 }}^{ - }+6{ H }^{ + }+S{{ O }_{ 3 }}^{ 2- }\longrightarrow $$

Explain the following:
$$(i)$$ Why do transition metal ions possess a great tendency to form complexes?
$$(ii)$$ The paramagnetic character in $$3d$$-transition series elements increases upto $$Mn$$ and then decreases.

Give the composition of the following alloys:
(a) German silver
(b) Brass

Explain:
(i) Transition elements mostly form the complex compound. Why?
(ii) Transition elements are good catalyst. Why?
(iii) Transition elements show variable valency. Why?

Explain the following
(i) $$Cu^+$$ is colourless and $$cu^{2+}$$ is coloured,
(ii) $$Zn$$ shows only + 2 Oxidation state.

With reference to the first row transition series :
i) Name the metal which possesses maximum number of oxidation states.
ii) Among $$Zn^{+2}$$ and $$Cu^{+2}$$ which is colourless?
iii) Between $$Ti^2+$$ and $$V^{2+}$$ which ion contains more number of unpaired electrons?

(a) What are Lanthanide elements?
(b) What do you understand by Lanthanide contraction?
(c) Write chromyl chloride test with equation.

(a) Why Transition elements become paramagnetic?
(b) Explain oxidization properties of potassium permanganate in acidic medium. (any four only)

(i) Scandium (z = 21) is a transition element but zink (z = 30) is not. Explain.
(ii) Calculate the equivalent weight of $$KMnO_4$$in acidic medium.
(iii) What do you mean by Lanthanoid contraction ?

Explain why:
Zr and Hf exhibit similar properties.
(At. no. of Zr = 40, Hf = 72)

(i) Write down any three difference between Lanthanoid and Actinoid ?
(ii) The melting and boiling points of Zy. Cd and Hg are low. Why ?
(iii) Draw the structure of manganate ion.

Explain why:
Transition elements usually form coloured ions.

Which of the following ions is coloured in aqueous solution?
i) $$Cu^{2+}$$
ii) $$Cu^{1+}$$

Transition metals and their compounds are used as catalysts.
Give two reasons.

Account for the following:
(i) Manganese shows maximum number of oxidation states in 3d series.
(ii) $$E^o$$ value for $$Mn^{3+} / Mn^{2+}$$ couple is much more positive than that for $$Cr^{3+} / Cr^{2+}$$.
(iii) $$Ti^{4+}$$ is colourless whereas $$V^{4+}$$ is coloured in an aqueous solution.

Fill in the blanks:
(N) Vegetable ghee is manufactured by the ..... of oils in presence of ..... as a catlayst.

Give reason :
Transition metals and their compounds show catalytic activities.

Write the preparation of the following :
$$Na_2 Cr_2 O_7$$ from $$Na_2CrO_4$$

$$NiCl_2$$ in the presence of dimethyl glycoxime (DMG) gives a complex which precipitates in the presence of $$NH_4OH$$, giving a bright red color.
Predict whether it is paramagnetic or diamagnetic.

Predict the spin only magnetic moment for:
$$Mn^{2+}$$

Predict the spin only magnetic moment for:
$$Fe^{2+}$$

Answer the following:
Why the compounds of transition elements are coloured?

Explain the following giving proper reason:

Transition metals and their compounds act as catalysis.

Explain the following giving proper reason:

$$CuSO_4$$ is paramagnetic while $$ZnSO_4$$ is diamagnetic.

Predict the spin only magnetic moment for:
$$Cu^{+}$$

Explain by giving suitable reason.

Yellow coloured aqueous solution of sodium chromate changes to orange-red when $$CO_2$$ under pressure is passed.

Predict the spin only magnetic moment for:
$$Cr^{6+}$$

Answer the following:
Why most of the compounds of transition elements are paramagnetic?

Explain by giving suitable reason.

$$Hg^{2+}$$ and $$Hg_2^{2+}$$ salts are colourless.

Answer the following:
Which bivalent ion of first transition series show the highest paramagnetism?

Explain:
Which divalent metal ion has maximum paramagnetic character among the first transition metals? Why?

Explain by giving suitable reason.

$$Cu^{2+}$$ salts are paramagnetic while $$Cu^{+}$$ salts are diamagnetic in nature.

Write down the name of catalyst for the following:
Reppe synthesis of benzene.

Write down the name of catalyst for the following:
Fenton's reagent in the synthesis of aldehydes from alcohols.

Write down the name of catalyst for the following:
Deacon's process of making $$Cl_2$$ from $$HCl$$.

Write down the name of catalyst for the following:
Adams catalyst in reduction.

Arrange following complex in decreasing order of magnetic moment:

$$[Ni(H_2O)_4]^{2+},\ [Ni(CN)_4]^{2-},\ [Fe(CN)_6]^{3-},\ [Fe(CN)_6]^{4-}$$

What happen when?
Ammonium dichromate is heated strongly.

Complete and balance the following reaction:

$$Cr_{2}O_{7}^{2-} + I^{-} + H^{+} \rightarrow$$