Class 11 Medical Chemistry - The P-Block Elements

Write two point of dissimilarities between boron and aluminium.

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Match List I with List II.

The list I contain formulae of ions and List II contains their name.

Bicarbonate ion = $$HC{ O }_3^-$$

Sulphate = $$SO_4^{2-}$$

Bisulphate = $$HS{ O }_4^-$$

Phosphate = $$PO_4^{3-}$$

Hypochlorite = $$ClO^-$$

Hence the answer is $$A\ = 5$$; $$B\ = 4$$; $$C\ = 3$$; $$D\ = 2$$; $$E\ = 1$$.

An element X is placed in groupState the formula and the nature of bonding in its chloride. Draw its electron dot structure of its chloride.

As the element X is placed in group 14 then the valency of the element is 4.

Now we know that the valency of chlorine is 1 as it's the atomic number is 17.

So as the element X has the valency 4 it can neither gain nor lose electrons to attain octet.

X is probably C.

So it makes a covalent bonding with four atoms of chlorine to form a compound $$CCl_4$$.

If true enter 1, else enter 0.
Although both carbon and silicon are in group IVA(14), very few Si - Si bonds are known.

Although both carbon and silicon are in group IVA(14), very few Si-Si bonds are known. This is because of larger size of silicon (Si) results in poor sideway overlap of $$p$$ -orbitals to form $$\pi$$-bond as compared to cabon atoms.

Carbon is an essential element in the molecule on which life is based but silicon is only satisfactory.
If true enter 1, else enter 0.

Carbon forms various stable compounds with itself and other elements such as $$N.O,H,S.$$ Bimolecules contain chains and rings with many $$C-C$$ bonds. $$Si-Si$$ bonds are much less stable and chain of $$Si$$ atoms are uncommon.

Therefore, the given statement is true.

The oxidation state of carbon in ionic carbide is $$x$$. The value of $$x\:+\:10$$ is :

$$Be_2C$$ and $$Al_4C_3$$ are ionic carbides; oxidation state of carbon $$=-4$$

$$x = -4$$

$$x+ 10 $$ = $$( -4)+ 10$$

$$x$$ = $$6$$

Match the compounds given in list I with the oxidation state of their carbon given in list II.

(A) In CO, the oxidation state of carbon is $$+2$$ as oxygen has $$-2$$ and so, the compound is overall neutral.

(B) In HCN, the oxidation state of carbon is $$+2$$ as nitrogen is in -3 oxidation state and H in $$+1$$.

(D) In $$CaC_2$$, the oxidation state of carbon is $$-2$$ as $$Ca^{2+}$$ has valency of $$+2$$.

State whether the following statement is true or false:

Tetra covalent nature of the element with following valence shell electronic configuration is : $$1s^2$$ $${2s}^2$$ $${2p}^2$$ explain on the basis of excitation of $$2s$$-electrons to $$2p$$-orbital.

The electronic configuration of Carbon is $${1s}^2$$ $${2s}^2$$ $${2p}^2$$

Since one sub-shell of 2p shell is vacant, one electron jumps from 2s shell to 2p shell creating a covalency of 4. Hence , it is tetra covalent.

Therefore, statement is true.

Give the formula of a compound (given in list II) in which carbon exhibits an oxidation state of (given in list I):

$$(a)$$ $$+4$$ oxidation state is shown by $$CO_2$$ and $$CCl_4$$.

As O and Cl has valency of $$-2$$ and $$-1$$ respectively.

$$(b)$$ $$+2$$ oxidation state is shown by $$CO$$.

As O has valency of $$-2$$.

$$(c)$$ $$-4$$ oxidation state is shown by $$CH_4$$.

As H has valency of $$+1$$.

Out of the elements $$C, S, Ge, Sn$$ and $$Pb$$, only $$+4$$ oxidation state is shown by how many element(s)?

C shows only +4 oxidation state.
Sn, Ge and Pb shows both +2 and +4.
S shows many oxidation numbers including 4.

Hence, the answer is 1.

How many elements in group $$14$$ are used as semiconductor?

Si and Ge are include in carbon family ( 14- group ). Both have four valence electrons, but germanium at a given temperature have more free electrons and a higher conductivity. Silicon is used as a semiconductor for electronics,

Therefore, 2 elements in group-14 are used as semiconductor.

Write the equation for the action of heat on a aluminium hydroxide.

$$2NaAlO_2 + CO_2 +3H_2O \rightarrow 2Al (OH )_3 + Na_2CO_3$$

$$2Al(OH)_3 \rightarrow Al_2O_3 + 3H_2O$$

Inert pair effect is shown by how many elements?

$$Ga, Al, Tl, Pb, Sn, Ge$$

Inert pair effect is shown by five elements $$Ga, Tl, Pb, Sn, Ge$$ It is not shown by Al.

The inert pair effect is the tendency to form ions two units lower in charge than expected from the group number of heavier $$p$$ - block elements.

If the statement is true enter 1 else enter 0.

The inert pair effect is the tendency to form ions two units lower in charge than expected from the group number of heavier p-block elements.

For example in group IVA, the +2 oxidation state of Pb is more stable than +4 oxidation state as 6s electrons do not participate in the ionization process and remain as inert pair.

Draw the structure of following compound.
$$[B_3O_3(OH)_4]^-$$

The structure of $$[B_3O_3(OH)_4]^-$$ is shown in the image.

$$SiCl_4\xrightarrow {H_2O}X\xrightarrow {1000}Y$$
In the above reaction, how many oxygen atom(s) is/are present in X and Y ?
[If answer is 6 and 4, represent as 64]

$$SiCl_4 \xrightarrow {H_2O} \underset {X} {Si(OH)_4} \xrightarrow {1000} \underset {Y} {SiO_2}$$
In the above reaction, four oxygen atoms are present in X and two oxygen atoms are present in Y. Hence, the answer is 42.

In silica, $$SiO_2$$, each silicon atom is bonded to how many oxygen atom?

In silica, $$SiO_2$$, each silicon atom is bonded to four oxygen atoms. $$SiO_2$$ unit forms a regular tetrahedron with Si at center and 4 O atoms at four corners.

Draw the structure of following compound.

$$Na[B_4O_5(OH)_4].8H_2O$$

$$Na[B_4O_5(OH)_4].8H_2O$$ (Borax)

Match the following and select the correct option:

A. Human body metabolism is the breakdown of food into the constituent particles is a slow process of combustion.

B. Fire crackers contain explosive material which explodes quickly.

C. Burning of domestic gas is a rapid process as it contains ethane,propane,butane,pentane.

D. Yellow phosphorous burns as it is flammable and pyrophoric (self igniting) upon contact with air and odour of this conbustion is garlic like.

Sulphuric acid formula is [$$H_{2}SO_{3}$$]. If this is true enter 1 , if false enter 0.

Silicon is the most abundant element on earth
If this is true enter 1, if false enter 0.

On earth, oxygen is the most common element, making up about 47% of the earth's mass. Silicon is second, making up 28%, followed by aluminum (8%), iron (5%), magnesium (2%), calcium (4%), sodium (3%), and potassium (3%). All of the remaining elements together make up less than 1% of the earth's mass.

Explain why is there a phenomenal decrease in ionization enthalpy from carbon to silicon?

The atomic radii of C and Si are 77 and 118 pm respectively.
Thus, the atomic radius increases significantly from C to Si. This increase is gradual from Si to Ge.
Higher is the atomic radius lesser is the energy required to release electron and lesser is the ionization energy.
Hence, there a phenomenal decrease in ionization enthalpy from carbon to silicon.

How would you account the lower atomic radius of $$Ga$$ as compared to $$Al$$.

The lower atomic radius of Ga as compared to Al can be explained on the basis of the presence of 3d series of transition metals. The size of d orbitals is larger than that of p orbitals. This results in poor shielding in d block elements. Hence, the effective nuclear charge of Ga is lower than the expected value. This results in lower atomic radius.

$$BF_3$$ dimerizes to form stable molecule.
Enter 1 if true else enter 0

Answer 0

$$BF_3$$ do not form Dimer.

Boron trifluoride is the inorganic compound with the formula $$BF_3$$. $$BF_3$$ is Lewis acid which is electron deficient and shows exothermic reactivity toward Lewis bases. When Boron forms bonds with fluorine, boron has $$6$$ electrons and fluorine has $$8$$ electrons. Now fluorine being small in Size and having High Electronegativity (tendency to pull negative charge) has a high amount of charge density (charge per volume) on it, which it prefers to reduce. So, it loses an electron and Boron accepts that electron. Then Fluorine makes a bond with the electron it just lost, forming a double bond. This is known as $$p(\pi)-p(\pi)$$ back bonding. As a result of this, both boron and fluorine have eight electrons. All three fluorine atoms are capable of this type of bonding, making boron a highly stable molecule.

While Borane $$BH_3$$ dimerizes to form stable molecule diborane $$B_2H_6$$.

Discuss the pattern of variation in the oxidation states of (i) $$B$$ to $$Ti$$ and (ii) $$C$$ to $$Pb$$.

(i) B shows $$+3$$ oxidation state. Other group 13 elements s show $$+1 $$ and $$+3$$ oxidation state.
On moving down the group 13, the stability of $$+1$$ oxidation state increases and the stability of $$+3$$ oxidation state decreases due to inert pair effect.
B and Al show $$+3$$ oxidation state, $$Ga$$ and $$In$$ show $$+1$$ and $$+3 $$ oxidation states and $$Tl$$ mainly shows $$+1$$ oxidation state.

(ii) Group 14 elements show $$+2$$ and $$+4$$ oxidation states. The stability of $$+2$$ oxidation state increases and the stability of $$+4$$ oxidation state decreases due to inert pair effect. $$C$$ and $$Si$$ mainly show $$+4$$ oxidation state, $$Ge$$ and $$Sn$$ show $$+2$$ and $$+4$$ oxidation state and $$Pb$$ mainly shows $$+2$$ oxidation state.

Why is $$B_2$$ paramagnetic whereas $$C_2$$ diamagnetic?

For $$B_2(10_{e^-})$$, the MO configuration is $$(\sigma1s)^2(\sigma^{\star}1s)^2(\sigma2s)^2(\sigma^{\star}2s)^2(\pi2p^1_x\approx\pi2p^1_y)$$

Due to the presence of unpaired electrons $$(2p^1_x=\pi2p^1_y)B_2$$ shows paramagnetism.

For $$C_2(12e^-)$$, the MO configuration is $$(\sigma1s)^2(\sigma^{\star}1s)^2(\sigma2s)^2(\sigma^{\star}2s)^2(\pi2p^{2_x}\approx.\pi2p^{2_y})$$

No unpaired electrons are there in $$C_2(\pi2p^2_x=\pi2p^2_y)$$, hence it shows diamagnetism.

The valencies of p block elements are equal to group number or 8 minus the group number. Therefore they show one lower valency and one higher valency. However, if we go to the bottom elements of any group, the compounds with lower valency are predominant whereas higher valency becomes rather uncommon. Explain.

It is due to Inert pair effect.It is increasing stability of oxidation states that are two less than the group valency for the heavier elements of groups 13,14 ,15 and 16.

Consider as an example thallium (Tl) in group 13 . The +1 oxidation state of Tl is the most stable, while Tl^III compounds are comparatively rare. The stability of the +1 oxidation state increases in the following sequence:

Al^I < Ga^I < In^I < Tl^I.

Describe an activity which provides the evidence for
a) the motion of particles
b) attraction between particles
c) inter-particle space

(a) Activity to show evidence for motion of particles: Take two glass beakers. Fill one beaker with hot water and another with cold water. Add $$KMnO_4$$ crystals in very small amount in both. Hence we observe the motion of particles that , the particles of $$KMnO_4$$move fastly in hot water than in cold water.

(b) Activity to show evidence for attraction between particles: Take a piece of iron and a piece of chalk. Try to break them. We come to know that iron is too hard to break. From this , we say that the particles of matter have forces acting between them. That force kept the particles together. The force of attraction between particles is maximum in solids and minimum in gases.

(c) Activity to show evidence for inter particle space: Take a glass beaker. Fill it with water. Measure the level of water with a scale. Add a tea spoon of salt and stir it well. We observe that there is no change in the water level. We can conclude that there exists inter particle space in the substances like water. Inter particle space is more in gases and less in solids.

What is inert pair effect?

Explanation:

Inert pair effect can be defined as the reluctance of the outermost shell $$s-$$ electrons to participate in bonding.

It is most commonly observed for group $$4$$ and group $$5$$ elements.

The lower oxidation states of the elements of this group are formed due to the inert pair effect.

Example: For $$Sn$$, there are two oxidation states $$+2$$ and $$+4$$. Out of these, $$Sn^{2+}$$ is formed due to the inert pair effect, where the valence shell $$s-$$ electrons remain paired instead of participating in bond formation.

Write the general valence shell electronic configuration of p-block elements.

The general valence shell electronic configuration of $$p-$$ block elements is $$ns^2np^{1-6}$$. Except Helium, all the other elements in $$p-$$ block follow this configuration.

Based on M.O.T, explain the electronic configuration of the Hydrogen molecule and $$B{e_2}$$.

Hydrogen molecule($$H_2$$) :- 2 electrons

electronic configuration :- $$\sigma 1s^2$$

It is diamagnetic because it has no unpaired electons

$$Be_2$$ :- 8 electrons

Electronic configuration :- $$ \sigma{1s}^2\sigma^*1s^2\sigma2s^2\sigma^*2s^2$$

It is also dimagnetic as it has no unpaird electrons.

Draw electron dot representation for the formation of magnesium chloride.

[Mg=12, Cl=17]

Magnesium donates one electron each with two chlorine atoms resulting in the formation of magnesium chloride.

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Why does boron not form $${B}^{3+}$$ ion?

As the Boron atom is small in size a large amount of energy is needed to remove 3 electrons from the boron atom.

The atomic number of Boron is 5. Its electronic configuration is $$1s^22s^22p^1$$.

When one electron is
removed from the p orbital a He-like fulfilled s orbital is left. This is
highly stable. So the second ionization enthalpy is quite high.

Again when one electron is removed a half filled orbital is left. So the third ionization enthalpy is also quite high.

Since the total energy needed to make $$ B^{3+} $$is the total of all the ionization enthalpies an enormous amount of energy is required to form it.

Name one metalloid of period 3.

A metalloid is a chemical element that exhibits some properties of metals and some properties of nonmetals.

One metalloid of period 3 is Silicon.

Carban and silicon are mainly tetravalent but $$Ge,Sn$$ and $$Pb$$ show divalency.

Carbon (C), Silicon (Si), Geranium (Ge), Stannum (Sn/Tin), and Plumbum (Pb/Lead) are all elements that belong to the p block of the periodic table. They have an electronic configuration of the elements is $$ns^2 np^2$$.
Of the p-block elements, Ge, Sn, and Pb are known to be divalent most of the time, when compared to C and Si, which are tetravalent most of the time.
This is because Ge, Sn, and Pb have an inert pair effect. The inner s electrons have inertness due to the poor shielding capacity of the outer electrons, which brings down the oxidation state of the elements by 2.The valence p-electrons are readily available for bonding while the s-electrons are attached to the nucleus.
Therefore, the elements with large atomic numbers are divalent and C and Si are tetravalent.

$$CO_{2}$$ is gas, while $$SiO_{2}$$ is a solid at room temperatue. State a reason for it.

$$CO_2$$ consists of individual molecules with one central carbon atom double bonded to two oxygen atoms. Silicon does not form double bonds with oxygen.$$CO_2$$ is gas but $$SiO_2$$ is solid because the $$SiO_2$$ is in giant molecular structure but the $$CO_2$$ is simple molecular structure.Because of this reason $$CO_2$$ is gas but $$SiO_2$$ is solid.

Why the compounds of $$Tl^+$$ are much more stable than those of $$Tl^{3 +}$$ ?

$$Tl = ns^2 np^1$$
$$Tl^{3+} = ns^onp^o$$
But $$Tl^+ = ns^2 , np^o$$
This metal gas d & f orbital which has poor shielding due to which $$ns^2 \, e^-$$ pair does not participate in bonding.

Mention the variation of atomic radii among the group 13 elements. Explain why ?

The correct order of atomic radii in group 13 elements is

B < Al < Ga < In =Tl

The atomic radii increase down the group but the increase is very less at the bottom of the series.

This is because of the screening effect.

Which type of bond is formed in $$BF_{3}$$?

$$BF_3$$ is covalent bond.

Since they mutually share their electrons.

Bond is formed due to high ionisation energy of boron.

What is the reduction power of the hydrides of group 14 elements in increases order?

Reduction power of the hydridies of group 14 elements increase in the order

$$CH_4 < SiH_4 < GeH_4 < SnH_4 < PbH_4$$

Because, $$Si-H$$ bond weaker than $$C-H$$ bond due to more bond length of $$Si-H$$ bond.

Similarly, $$GeH_4 (Ge-x)$$ is weaker than $$Si-H$$ and $$C-H$$ bond

Similarly, $$Sn-H$$ is weaker than $$Ge-H, Si-H, C-H$$

Similarly, $$Pb - H$$ is weaker than $$Sn - H, Ge-H, Si-H, C-H$$

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Write about the nature of carbon atom

Carbon has the atomic number of 6, meaning each carbon atom has a total of six electrons.

Small size of the carbon atom makes the compounds of Carbon exceptionally stable.

Hence carbon as an element has the ability to form a variety of stable compounds, which can exist freely in nature.

The aluminium metal is obtained by the reduction of aluminium oxide with:

Bauxite is the aluminium ore. It is purified to yield aluminium oxide, which is a white powder. Aluminium is then extracted from aluminium oxide by electrolysis.The aluminium oxide must be made molten so that the ions can pass through it during electrolysis.
The aluminium metal is obtained by the reduction of aluminium oxide with Carbon.

What are interhalogen compounds? Explain with two examples.

The inter-halogen compound is a molecule that contains two or more different halogen atoms (fluorine, chloride, bromide, iodine as astatine) and no atoms of other groups.

Most interhalogen compounds are known as binary (composed of only two distinct elements). Their formulas are generally $${ XY }_{ n }$$, where $$n=1,3,5$$ or $$7$$, and $$X$$ is less electronegative than $$Y$$.

Examples are $${ Br }_{ 2 },{ Cl }_{ 2 },BrCl,{ \left( I{ Cl }_{ 3 } \right) }_{ 2 }$$ etc.

The element with atomic number 36 belongs to______ block in the periodic table.

Electronic configuration of atomic no. $$36$$ is

$$1s^22s^22p^63s^23p^63d^{10}4s^24p^6$$

Last electron is present in p-subshell so this element is present in p-block.

Action of heat on aluminium hydroxide.

Aluminium hydroxide on thermal decomposition produce Aluminium oxide and water. The reaction takes place at $$575^o$$ $$C$$

$$2Al(OH)_3\longrightarrow Al_2O_3+3H_2O$$

Which of the following is the most basic : $$Al_2O, Tl_2O_3, Tl_2O.$$

$$Tl_2O$$ is more basic in nature compared to $$Al_2O_3$$ and $$Tl_2O_3$$ because farther down the table, lower the oxidation state, more basic is the oxide. Here $$Tl$$ is the bottom element of the table and has least oxidation state.

Account for the following
The tendency to form p$$\pi $$-p$$\pi $$ bonds of carbon is strong. Why $$Pb{ O }_{ 2 }$$ is a strong oxidizing agent than $$Sn{ O }_{ 2 }$$.

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Fill in the blanks:
__________ is present in both living and non-living things.

Carbon is present in both living and non-living things.

How is boron absorbed by plants?

Boron is absorbed as $$BO_{3}^{-3}$$ or $$B_{4}O_{7}^{2-}$$.

Give reasons for the following three statements.
a) Al is resistant to air and water
b) Boron halides are covalent in nature
c) $$PbCl_2$$ is more stable than $$PbCl_4$$

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Write four characteristic properties of p - block element.

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Write the general electronic configuration of s, p, d and f-block elements.

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Fill in the blank:

Compounds that formally contain $$Pb^{4+}$$ are easily reduced to $$Pb^{2+}$$. The stability of the lower oxidation state is due to __________.

Compounds that formally contain $$Pb^{4+}$$ are easily reduced to $$Pb^{2+}$$. The stability of the lower oxidation state is due to $$\text{inert pair effect.}$$

In the periodic table, inert effect increase down the group. So in group $$14$$, the stability of smaller oxidation state of $$+2$$ goes on increasing down the group. $$Pb^{4+}$$ has an oxidizing nature and $$Br^-$$ and $$I^-$$ have reducing nature, so $$Pb^{4+}$$ is reduced to $$Pb^{2+}$$

$$Pb^{4+}+2Br^- \rightarrow Pb^{2+}+Br_2$$

$$Pb^{4+}+2I^-\rightarrow Pb^{2+}+I_2$$

Does Ga has $$+2$$ oxidation state in $$GaCl_2$$?

GaGI2GaGI2 is actually

Ga2CI4Ga2CI4, i.e.,

Ga+[Ga3+CI4]Ga+[Ga3+CI4]. Ga has thus

+1+1 and

+3+3 oxidation states and not

+2+2.

Answer the following
Identify the most abundant metal in the earth's crust and enter the group number it belongs to.

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Among group $$14$$ elements name:
The element having highest tendency to from $$p \pi -p \pi$$ bonds.

Due to small size and higher electronegativity, Carbon has the highest tendency to form multiple bonds either with itself or with the other atoms of similar size such as oxygen and nitrogen.

Lead prefers to form divalent compounds. Give reason.

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Give reason for each of the following:
Why $$Ge, Sn$$ and $$Pb$$ show divalency?

Of the p-block elements, Ge, Sn, and Pb are known to be divalent most of the time, when compared to C and Si, which are tetravalent most of the time. This is because Ge, Sn and Pb have an inert pair effect. The inner s electrons have inertness due to the poor shielding capacity of the outer electrons, which brings down the oxidation state of the elements by 2.The valence p-electrons are readily available for bonding while the s-electrons are attached to the nucleus.

Among group $$14$$ elements name:
The element which is commonly found in $$+2$$ oxidation state.

Since the inert pair effect becomes more and more predominant down the group, therefore, stability of +2 oxidation state increases down the group. thus, Lead is commonly found in +2 oxidation state.

Answer the following
Name the element of group $$13$$ which forms the most stable compounds in $$+1$$ oxidation state.

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Answer the following
What is the outer electronic configuration of group IIIA elements?

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Explain the following with proper reasoning.
$$PbO$$ does not dissolve in $$H_2SO_4$$ while $$SnO$$ is soluble in $$H_2SO_4$$.

The surface of metallic lead is protected by a thin layer of lead oxide $$PbO$$. This renders the lead essentially insoluble in sulfuric acid.

But in case of Tin, $$SnO$$ does not protect the surface of metallic tin.

So, $$PbO$$ does not dissolve in $$H_2SO_4$$ while $$SnO$$ is soluble in $$H_2SO_4$$.

What is the oxidation state of iodine in $$ H_5IO_6? $$

Let oxidation state of iodine be $$'x'$$

$$ 5(+1) + x -6(-2) = 0 $$

$$x = +7 $$

The oxidation state of iodine is $$+7$$

Explain the following with relevant reason.
Although the ionisation potential of boron $$(8.30 eV)$$ is less than gold $$(9.22 eV)$$, yet former is a non-metal while the latter is a metal.

This is due to structural difference in the solid states. In general, metals have large number of atoms as neighbours while non-metals have relatively less number of atoms as neighbours. Gold has $$12$$ atoms as neighbours while boron has $$6$$ or less atoms as neighbours in solid states.

Three pairs of compounds are given below. Identify that compound in each of the pairs which has group 13 element in more stable oxidation state. Give reason for your choice. State the nature of bonding also
(a)$$TlCl_3$$, $$TlCl$$
(b)$$AlCl_3$$, $$AlCl$$
(c)$$InCl_3$$, $$InCl$$

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Describe the general trends in the following properties of the element in group 13 and 14
(a) Atomic size
(b)Ionization enthalpy
(c)Metallic character
(d)Oxidation states
(e)Nature of halides

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In group $$15$$ elements, the number of unpaired electrons in valence shell is:

Correct answer :- 3

The general electronic configuration of 15th group element is ns2 np3.Hence, total unpaired electrons are 3

Identify the compounds A, X, and Z in the following reactions:
(a) $$A+2HCl +5H_2O \rightarrow 2NaCl +X$$

$$X\xrightarrow [ 370K ]
{ \Delta } HBO_2\xrightarrow [>370K] {\Delta}Z$$

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Compare the properties of boron and aluminium under the following heads
Electronic configuration

Boron configuration is $$[He] 2s^22p^1$$

and Aluminium configuration is $$[Ne]3s^23p^1$$

What happens when water reacts with the following compound?

Borax

$$Borax$$ also known as sodium tetraborate or sodium borate, is a mineral and inorganic salt. Borax can be found as anhydrous, pentahydrate or decahydrated salt.

Borax in water form alkaline solutions.

Aqueous solution of borax acts as a buffer because it contains weak acid and its salt with strong base.

$$Na_2B_4O_7 + 7H_2O\rightarrow 2Na[B(OH)_4] + 2H_3BO_3$$

$$Borax$$ is sparingly soluble in cold water but fairly soluble in hot water. Borax reacts with water to form sodium hydroxide ($$NaOH$$), a strong base and boric acid ($$H_3BO_3$$), a weak acid.

$$Na_2B_4O_7+7H_2O\space \rightarrow2NaOH +4H_3BO_3$$

The number of unpaired electrons in the valence shell of the members of oxygen family is:

Valence shell electronic configuration of members of oxygen family is $$ns^2 np^4.$$

in the following manner
$$\overset { Valence\quad shell }{ { \overset { ns }{ \boxed { \upharpoonleft \downharpoonright } } }{ \quad \overset { np_{ x } }{ \boxed { \upharpoonleft \downharpoonright } } }\overset { np_{ y } }{ \boxed { \upharpoonleft } } \overset { np_{ z } }{ \boxed { \upharpoonleft } } } $$

$$\therefore$$ two electrons are unpaired in the valnece shell of member of oxygen family.

Compare the properties of boron and aluminium under the following heads
Nature of chlorides

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Compare the properties of boron and aluminium under the following heads:
Nature of oxides

The oxides and hydroxides of boron are acidic in nature whereas, the oxides and hydroxides of aluminium are amphoteric in nature.

Name the various forms of carbon found on the Earth. How the carbon cycle may get disrupted.

On earth oceans are the largest reserves of carbon, after oceans forests are the second largest reserves.

Carbon can be found in its combined from as gas, $$ CO_{2} $$.
Elemental state of carbon is in form of compounds like diamond and graphite.
All life supporting molecules like, proteins,fats, nucleic acids etc. are made of carbon.
Process of photosynthesis i.e. formation of complex sugars from carbon di oxide and water in the presence of sunlight cannot be completed without carbon.
Carbon is present in soil as minerals, it forms the exoskeleton of a number of organisms.
Carbonates, coal reefs, animal shells everything inside oceans consists carbon.

Account for the following:
a. Ozone is thermodynamically unstable.
b. Solid $$PCl_{5}$$ is ionic in nature.
c. Fluorine forms only one oxoacid $$HOF$$.

(a) Ozone is thermodynamically unstable with respect to oxygen as its decomposition into oxygen results in the liberation of heat $$(DH$$ is $$-ve)$$ and an increase in entropy $$(DS$$ is $$+ve)$$. These two effects reinforce each other, resulting in large negative Gibbs energy change $$(DG)$$ for its conversion into oxygen.

(b) In solid state, $$PCl_{5}$$ exists as $$[PCl_{4}]^{+}[PCl_{6}]^{-}$$ and conducts electricity on melting.

How would you account for the following:
Both $$O_{2}$$ and $$F_{2}$$ stabilise higher oxidation states but the ability of oxygen to stabilise the higher oxidation state exceeds that of fluorine.

This is due to tendency of oxygen to form multiple bonds with metal atom.

What happens when
a. Chlorine gas is passed through a hot concentrated solution of $$NaOH$$?
b. Sulphur dioxide gas is passed through an aqueous solution of $$Fe(III)$$ salt?

a. $$3Cl_{2} + \underset {(Hot\ and\ conc.)}{6NaOH} \rightarrow 5NaCl + NaClO_{3} + 3HN_{2}O$$

b. $$2Fe^{3+} + SO_{2} + 2H_{2}O \rightarrow 2Fe^{2+} + SO_{4}^{2-} + 4H^{+}$$.

Account for the following:
a. $$SnCl_{4}$$ is more covalent than $$SnCl_{2}$$.
b. Tendency to form pentahalides decreases down the group in group $$15$$ of the periodic table.

(a) The oxidation states of central atom $$Sn$$ in $$SnCl_{4}$$ and $$SnCl_{2}$$ are $$+4$$ and $$+2$$ respectively. $$+4$$ state of $$Sn$$ has higher polarising power which, inturn, increase the covalent character of bond formed between the central atom and the other atoms.

(b) This is due to inert pair effect. The stability of $$+5$$ oxidation state decreases down the group in group $$15$$.

Explain the following observations:
a. Oxygen is a gas but sulphur a solid.
b. Despite having greater polarity, hydrogen fluoride boils at a lower temperature than water.
c. The halogens are coloured.

(a) Because of its small size, oxygen is capable of forming $$\rho \pi - \rho \pi$$ bond and exists as diatomic $$O_{2}$$ molecule. The intermolecular forces in oxygen are weak van der Waals for, due to which it is a gas at room temperature. On the other hand, sulphur, due to its larger size prefers to form $$S-S$$ single bond and exist as octaatomic $$S_{8}$$ molecule having puckered ring structure. Because of larger size the force of attraction holding the $$S_{8}$$ molecules together are much stronger. Hence sulphur is a solid at room temperature.

(b) This is because in $$H_{2}O$$ hydrogen bond formed is three dimensional whereas in $$H-F$$ it is linear.

(c) All halogens are coloured. This is due to absorption of radiation in visible region which results in the excitation of outer electrons to higher energy level while the remaining light is transmitted. The colour of the halogen is the colour of transmitted light.

Which of the following options is correct for the given statement:
Sand is a compound of
silicon and nitrogen
silicon and oxygen
oxygen and sulphur
none of the above

Sand i.e. silicon oxide ($$SiO_2$$) is a compound of silicon and oxygen.

How does carbon attain a stable electronic configuration ?

Carbon attains stable electronic configuration by sharing its four electrons with other atoms.

Explain each of the following:
a. Nitrogen is much less reactive than phosphorus.
b. The stability of $$+5$$ oxidation state decreases down group $$15$$.
c. The bond angles $$(O-N-O)$$ are not of the same value in $$NO_{2}^{-}$$ and $$NO_{2}^{+}$$.

a. As $$N\equiv N$$ triple bond $$(941.4\ kJ\ mol^{-1})$$ is much stronger than $$P-P$$ single bond $$(213\ kJ\ mol^{-1})$$, therefore nitrogen is much less reactive than phosphorus.

b. Due to inert pair effect stability of $$+5$$ oxidation decreases down the group $$15$$.

c. In $$NO_{2}^{-}$$, nitrogen has a lone pair-bond pair repulsion is greater than bond pair-bond pair repulsion, thus $$O-N-O$$ bond angle in $$NO_{2}^{-}$$ is less then $$NO_{2}^{+}$$.

Explain the properties of carbon.

Carbon has the following properties:

Physical properties:
$$\cdot$$ Carbon is a soft, dull gray or black non-metal.
$$\cdot$$ It exists in two forms i.e. crystalline and non crystalline forms. This property is termed as allotropy.

Chemical properties:
The chemical properties of carbon can be explained on the basis of reactions which carbon and its compound show.

1. Combustion reaction:
Carbon, in all its allotropic forms, burns in oxygen to give carbon dioxide along with the release of heat and light, This reaction is termed as combustion reaction.

The chemical equation to represent the burning of carbon is as follows:
$$\underset{Carbon}{C(s)} + \underset{Oxygen}{O_2(g)} \rightarrow \underset{Carbondioxide}{CO_2(g)} + Heat + Light$$

2. Oxidation reaction: Pure carbon is oxidized to give carbon dioxide when burnt in air. Similarly, hydrocarbons can be easily oxidized on burning. Substances that have the ability to oxidize other substances are known as oxidizing agents. Alkaline potassium permanganate and potassium dichromate are examples of oxidizing agents. For example, the oxidation of ethanol in the presence of potassium permanganate leads to the formation of ethanoic acid.

$$\underset{Ethanol}{CH_3 \,CH_2-OH} \xrightarrow[Alkaline KMnO_4]{\Delta} CH_3 - \underset{Ethanoic \,acid}{COOH}$$

3. Addition reaction: In an addition reaction, a substance is added to a hydrocarbon. A single product is obtained in this reaction. In unsaturated hydrocarbons, double or triple bonds are present. Atoms of different elements like hydrogen, chlorine, bromine etc. are added to these compounds across the double or triple bonds using different reagents and catalysts.

4. Substitution reaction: It is a type of reaction in which an atom or a group of atoms replaces another atom present in a molecule undergoing the reaction. For example, chlorine replaces hydrogen from methane in the presence of sunlight.

$$CH_4 + Cl_2 \overset{Sunlight}{\longrightarrow} CH_3Cl + HCl$$

Give different forms in which carbon occurs in nature.

Carbon occurs in free form as graphite and diamond. In combined form it exists as carbon dioxide, carbonates, etc.

Write the characteristics of p-block elements.

(i) p-block elements consists of metals, non-metals and metalloids.

(ii) p-block consists of solids, liquids and gases.

Fill in the blanks.
The name 'carbon' is derived from the Latin word __________.

The name 'carbon' is derived from the Latin word carbo.

Write name.
The oxide that forms salt and water by reacting with both acid and base.

Metal oxides which react with both acids as well as bases to produce salts and water are also known as amphoteric oxides. For example: $$Al_2O_3$$ is an amphoteric oxide. An amphoteric compound is a molecule or ion that can react both an acid as well as a base to produce salt and water.

What is tincal? Give its chemical formula.

Tincal (or Borax) is an ore of boron. Its formula is $$Na_2B_4O_7.10H_2O$$.

Why does boron form stable electron deficient compounds?

Boron has three valence electrons, it will share three electrons with other elements to form electron deficient compounds which are stable.

Why is boron metalloid?

Boron resembles with both metals and non-metals therefore, it is metalloid.

In which pure form does carbon exists in nature?

Fullerene is the purest form of carbon.

Name two allotropes of boron.

Boron can form $$\alpha$$-tetragonal, and $$\gamma$$-orthorhombic allotropes.

Name two isotopes of boron

$$_{5}^{10}$$ Band $$_{5}^{11}B$$ are isotopes of boron.

Complete the following
(i) $$ B + O_2 \rightarrow$$
(ii) $$ B + N_2 \rightarrow$$
(iii) $$ B + Cl_2\rightarrow$$
(iv) $$ BF_3 + NH_2\rightarrow$$
(v) $$ Na_2B_4O_7 + H_2O\rightarrow$$
(vi) $$Na_2B_4O_7\rightarrow$$

(i) $$ 4B + 3O_2 \xrightarrow\ 2B_2O_3$$

(ii) $$2B + N_2 \xrightarrow{heat} 2BN$$ (Boron nitride)

(iii) $$ 2B + 3Cl_2 \xrightarrow{\Delta} 2BCl_3$$ (Boron trichloride)

(iv) $$ BF_3 + NH_3 \rightarrow [F_3B\leftarrow NH_3]$$

(v) $$Na_2B_4O_7 + 7H_2O \xrightarrow 2NaOH + 4H_3BO_3$$

(vi) $$Na_2B_4O_7 \xrightarrow {heat} 2NaBO_2 + B_2O_3$$ (Borax bead)

Which type of bonds are formed by boron and why?

Boron forms covalent bonds because it cannot lose electrons or gain electron easily.

Why is crystalline boron hard solid?

It is due to strong covalent bonds.

Name the oxidation states exhibited by $$Ge, Sn$$ and $$Pb$$.

$$+2$$ and $$+4$$ are the most common oxidation states exhibited by $$Ge,Sn $$ and $$Pb$$.

Mention the state of hybridization of $$B$$ in $$BH_{4}^{-}$$

Hybridization of $$B$$ in $$BH_4^-$$ is $$sp^3$$.

What is state of hybridisation of $$C$$ in $$CO_{3}^{2-}$$?

$$C$$ in $$CO_{3}^{2-}$$ has $$sp^2$$ hybridisation.

What is name give to elements which are neither metals nor non-metals?

Metalloids, these are the elements which have the characteristics of the metals as well as the properties of the non-metals

Mention the chief reason for the anomalous behaviour of boron in group $$13$$ of the periodic table.

It is due to small size and higher ionization energy and high charge/size ratio of boron.

Write the state of hybridization in $$BF_3$$.

$$B$$ in $$BF_3$$ has $$sp^2$$ hybridisation

Carbon and Silicon are mainly tetravalent but $$Ge, Sn$$ and $$Pb$$ show divalency.
Give reason.

$$Ge, Sn$$ and $$Pb$$ are divalent due to inert pair effect which is not there in carbon and silicon.

Which one of the following elements $$+1$$ oxidation state as well: $$Al, B, Ca, Tl, Be$$?

$$Tl$$ shows $$+1$$ oxidation state due to inert pair effect.

What is the oxidation state of $$Ni$$ in $$[Ni(CO)_4]$$?

Oxidation state of $$Ni$$ is zero in $$[Ni(CO)_4]$$

Why does boron form electron deficient compounds?

It is because boron has three valence electrons. It shares three electrons and gets six electrons, i. e. its octet is not complete. Therefore, it forms electron deficient compounds

What name is given to the compounds formed by more electropositive elements with carbon?

A carbide usually describes a compound composed of carbon and a electropositive element (i.e. metal).

In terms of atomic size, ionization energy and charge, explain why are ionic compounds containing $$B^{3+}$$ not formed.

Boron has smaller atomic size, therefore, it cannot lose three electrons because its third ionization energy is very high.

Give the chemical reaction as an evidence for each of the following observation?
$$\cdot$$ Tin (II) is a reducing agent whereas lead (II) is not
$$\cdot$$ Gallium (I) undergoes disproportionation reaction.

$$\cdot$$ It is because $$Pb^{2+}$$ is more stable than $$Pb^{4+}$$ due to inert pair effect whereas $$Sn^{4+}$$ is more stable than $$Sn^{2+}$$. Therefore $$Sn^{2+}$$ is good reducing agent

$$\cdot$$ It is because $$Ga^{3+}$$ is more stable than $$Ga^+$$

State the trends observed in case of following:
Oxidation state of the elements of Group $$14$$.

Group $$14$$ elements show $$+2$$ and $$+4$$ oxidation states. Tendency to show $$+2$$ oxidation state increases down the group due to inert pair effect.

As we move down in group $$13$$ elements increase in atomic size is comparatively very less. Explain.

It is due to poor shielding effect of $$d$$ and $$f$$-electrons, as result of which effective nuclear charge increase.

Describe two similarities and two dissimilarities between $$B$$ and $$Al$$.

Similarities:

1. Both have similar electronic configuration

2. Both have same number of valence electrons

Dissimilarities:

1. $$B$$ is non-metal; $$Al$$ is metal

2. $$B$$ forms acidic oxide $$(B_2O_3)$$ whereas $$Al$$ form $$Al_2O_3$$ which is amphoteric

What is the oxidation state of $$C$$ in
(a) $$CO$$
(b) $$HCN$$
(c) $$H_2CO_3$$
(d) $$CaC_2$$.

(a) $$+2$$, (b) $$+2$$, (C) $$+4$$, (d) $$-1$$

What happens when a borax solution is acidified? Write a balanced equation for the reaction.

Boric acid is formed.

$$ Na_2B_4O_7 + 2HCl + 5H_2O \rightarrow 2NaCl + 4H_3BO_3$$

Arrange the following compounds in decreasing order of property indicated against each. Give reason for your answer :
$$BCl_3, AlCl_3, GaCl_3, InCl_3, TiCl_3$$ (Stability of $$+3$$ oxidation state)

$$BCl_3 > AlCl_3 > GaGl_3 > InCl_3 > TiCl_3$$ because of inert pair effect stability of $$+3$$ oxidation state decreases.

$$C$$ and $$Si$$ are always tetravalent but $$Ge, Sn, Pb$$ show divalency.

$$Ge, Sn\ and Pb$$ show divalency due to inert pair effect, $$Pb^{2+}$$ is more stable than $$Pb^{4+}$$.

Mention the nature of an aqueous solution of borax.

When borax is dissolved in water it gives $$NaOH$$ and boric acid. Since boric acid is a weak acid and $$NaOH$$ is strong a alkali an aqueous solution of borax is alkaline in nature

How does electron deficient compound $$BF_3$$ achieve electronic saturation i. e., the fully occupied outer electron shells?

1. By multiple bonding, e. g., $$BF_3$$ in which a lone pair of electrons of each fluorine atom may be used in a $$B\rightarrow F \ \pi $$ - bond (back bonding) involving the vacant orbital on the boron atom.

2. By formation of complexes in which electrons are received from a donor molecule, e.g., $$F_3B \leftarrow NH_3$$. Boron compounds, thus behave as Lewis acids.

Explain the oxidation property of group $$14$$ elements.

Group $$14$$ elements show commonly $$+2$$ and $$+4$$ oxidation state. The carbon family also shows both positive and negative oxidation state. It shows $$+2$$ oxidation state when $$np^2$$ electrons involved in the bond formation $$Ex-SnCl_2$$, $$PbSO_4$$. This show $$+4$$ oxidation state when $$ns^2np^2$$ electrons involved in the bond formation. $$Ex-SnCl_4$$

Give the difference in structure of the following pair of compounds : $$CO_2$$ and $$SiO_2$$.

$$CO_2$$ linear molecule and exists as monomer. It is gas while $$SiO_2$$ is solid at room temperature due to three dimensional network in which each $$Si$$ atom is covalently bonded to four oxygen atoms tetrahedrally. In $$CO_2$$, $$'C'$$ is sp-hybridised while in $$SiO_2$$, $$'Si'$$ is $$sp^3$$ hybridised. $$CO_2$$ is discrete molecule where as $$SiO_2$$ is $$3-dimensional$$ solid.

What is the electronic configuration of p-block elements ?

$$ ns^2np^{0 - 5} $$

What are allotropy? Mention allotropy of $$Pb, Sn, Ge, Si$$.

Allotropy: The existence of same element in two or more forms in the same physical state but having different properties called allotropy. The different forms are called allotrope.

$$Pb$$ : No allotrope exists.

$$Sn$$ : Tin exists in two different forms, or allotropes: the familiar form, white (or beta) tin, and gray (or alpha) tin.

$$Ge$$ : Two (Both are crystalline form.)

$$Si$$ : There are two allotropes of silicon: amorphous and crystalline

Compound $$X$$ on reduction with $$LiAH4$$ gives a hydride $$Y$$ containing $$21.72$$% hydrogen along with other products. The compound $$Y$$ reacts with air explosively resulting in boron trioxide. Identify $$X$$ and $$Y$$. Give balanced equations involved in the formation of $$Y$$ and its reaction with air. Draw the structure of $$Y$$.

Step 1 : To determine the molecular formula and structure of compound $$Y$$.

1. Since the hydride $$Y$$ reacts with air forming boron trioxide, therefore, $$Y$$ must be an hydride of boron.

2. %$$H = 21.72$$% (Given)

$$\therefore$$ Emperical formula of $$Y = BH_3$$

Since boron forms two types of hydrides, i.e., $$B_nH_{n+4}$$ (nidoboranes) and $$B_nH_{n+6}$$ (arachnoboranes), therefore, $$Y$$ must be a nidoborane with $$ n = 2$$.

Thus, $$M.F$$ of $$Y = B_2H_6$$. If $$Y$$ is $$B_2H_6$$ (diborane), then its structure must be as follows:

Bridges $$B...H = 134$$ pm Terminal $$ B - H = 119$$ pm

Step 2: To determine the structure of the compound $$X$$.

Since compound $$Y$$ i.e., $$B_2H_6$$ is formed by reduction of compound $$X$$ with $$LiAH_4$$,

therefore, $$X$$ must either $$BCl_3$$ or $$BF_3$$.

$$4BF_3 + 3LiAlH_3 \overset{either}{\rightarrow} 2B_2H_6 + 3LiAlF_3$$

$$X$$ $$Y$$

The equation representing the reaction of $$Y$$ with $$O2$$ may be written as follows :

$$B_2H_6 + 3O_2 \rightarrow 2B_2O_3 + 3H_2O$$

Diborane, $$Y$$

Thus $$X = BF_3$$ and $$Y = B_2H_6$$

How does metallic character vary among the group $$14$$ elements? How is it related to ionisation energy?

Carbon and silicon are non metals. Germanium is a metalloid and tin and lead are metals. Metallic character increases from carbon to lead. The change form non metallic to metallic character is due to the increase in atomic size from carbon to lead. Consequently the ionization potential decreases from carbon to lead.

Hence the electropositive character increases favouring ionic bonding especially with oxidation state $$+2$$. Whereas carbon and silicon show $$+4$$ oxidation state forming covalent bonding with electron sharing.

A compound $$'X'$$ or reduction with $$LiAH4$$ gives a hydride $$'Y'$$ that contains $$21.72$$% hydrogen along with other products. The compound $$'Y'$$ reacts with air explosively resulting in boron trioxide. Identify $$'X'$$ and $$'Y'$$. Give balanced reactions involved in the formation of $$'Y'$$ and its reactions with air. Draw the structure of $$Y$$.

$$'X'$$ is $$BCl_3$$ Boron Trichloride and $$Y$$ is $$B_2H_6$$ as percentage of hydrogen given is $$21.72$$%

Chemical reaction involved are

$$4BCl_3 + 3LiAlH_4 \overset{diborane}{\rightarrow} 3LiCl + 2B_2H_6 + 3AlCl_3$$

$$B_2H_6 + 3O_2 \rightarrow B_2O_3 + 3H_2O + heat$$

$$(Y)$$ Boron

Dilborane trioxide

1737235_1827190_ans_e04fa80f8c5e487b94d7b19eac3aaab0.png

Explain the oxidation state of carbon in $$Be_2 \ C$$ and $$Al_4 \ C_3$$.

Oxidation state of carbon in $$Be_2 \ C$$
Let oxidation state of carbon = $$x$$
$$\therefore \ \ 2(+2) + x = 0$$
$$4 + x = 0$$
$$x = - 4$$
$$\therefore$$ The oxidation state of carbon in $$Be_2 \ C = - 4$$
Oxidation state of carbon in $$Al_4 \ C_3$$
Let oxidation state of carbon = $$x$$
$$\therefore \ \ 4(+3) + 3x = 0$$
$$12 + 3x = 0$$
$$3x = -12$$
$$\therefore$$ The oxidation state of carbon in $$AI_4 \ C_3 = - 4$$

Write the general electronic configuration of p-block elements.

The general electronic configuration of p-block elements is $$ns^2np^{1-6}$$.

Explain the following:
(i) Catenation
(ii) Inert Pair Effect
(iii) Allotropy.

(i) Catenation—It is the linkage of atoms of the same element into longer chains. Catenation occurs most readily in carbon, which forms covalent bonds with other carbon atoms to form longer chains and structures. This property is due to strong C—C bonds.

(ii) Inert Pair Effect—The reluctance of s electrons in participating in chemical bonding. It is due to the poor shillding of d and f block electrons of inner shell which increases the effective nuclear charge.

The term 'inert pair' was first proposed by N. Sidgwick in 1927. The name suggests that the s electrons are more tightly bound to the nucleus and therefore more difficult to ionize.

For example, the p-block elements of the 4th, 5th and 6th period come after d-block elements but the electrons present in the intervening d- (and f-) orbitals do not effectively shield the s-electrons of the valence shell. Due to this, the inert pair of ns electrons remains more tightly held by the nucleus and therefore participates less in bonding.

(iii) Allotropy-Allotropy is the property of some chemical elements to exist in two or more different forms, in the same physical state, known as allotropes of these elements. Allotropes are different structural modifications of an element, the atoms of the elements are bonded together in a different manner.

For example, the allotropes of carbon include diamond, graphite and fullerenes.

Explain the various oxidation states found in group 14 elements.

The group 14 elements have four electrons in outermost shell and its electronic configuration is $$ns^2 \ np^2$$. They can form $$M^{4+}$$ or $$M^{4-}$$ ions to show ionic character or exhibit tetravalent covalent nature by sharing of four electrons in order to get stability. In the formation of $$M^{4+}$$ or $$M^{4-}$$ ions require huge amount of energy which is normally not possible. So, these elements do not form ionic bond rather they form covalent bonds in both oxidation states. Lead compounds in $$+2$$ state are stable and in $$+4$$ state are strong oxidising agents.

Why Boron forms electron deficient compounds?

Boron forms electron deficient compounds because of small size and high charge density of boron. Boron has very small atomic radius. Boron forms covalent compounds. It has 3 valence electrons which it prefers to share rather than donate as $$B^{3+}$$ ions is very small and unstable, so it forms sextet rather than octet. Due to incomplete octet of boron, it forms electron deficient compounds.

Analyze the chemical equation and find the reason for your observation.
$$C^0+2H_2{+1}S^{+6}O_4^{-2} \rightarrow C^{+4} O_2^{-2}+2H_2^{+1}O^{-2}+2S^{+4}O_2^{-2}$$
What is the oxidation state of element carbon?

The oxidation state of element carbon is Zero.

Analyze the chemical equation and find the reason for your observation.
$$C^0+2H_2{+1}S^{+6}O_4^{-2} \rightarrow C^{+4} O_2^{-2}+2H_2^{+1}O^{-2}+2S^{+4}O_2^{-2}$$
What about the carbon in carbon dioxide?

The oxidation state of carbon in carbon dioxides is $$+4$$

Why the melting point of Boron is unexpectedly high?

Boron has closely packed metallic structure. It also has small size. So, the melting point of Boron is unexpectedly high.

Write the stability of +1 oxidation state of group 13 elements in increasing order.

The increasing order of stability of +1 oxidation state of group 13 elements is given below : . Al < Ga < In < TI (due to inert pair effect).

How many electrons are there in the outermost shell of carbon?

Atomic Number of Carbon is 6

Electronic Configuration - 2,4

4 electrons in outermost shell

Find the position of carbon in periodic table and the complete the table.
Symbol .................
Atomic number .................
No. of electrons in the outermost shell .................
Valency .................
Metal/ non-metal? .................

Symbol	$$C$$
Atomic number	$$6$$
No. of electrons in the outermost shell	$$4$$
Valency	$$4$$
Metal/ non-metal?	Non-metal

Total number of Boron atoms in anionic part of borax which participate in back bonding.

Borax: $$2Na^{+} [B_{4}O_{5}(OH)_{4}]^{2-} . 8H_{2}O$$
$$sp^2$$-hybridized Boron atoms only participate in $$p\pi-p\pi$$ back bonding.

So, the no. of such boron atoms is 2.

Fill in the blanks given in list I with items in list II.

(a) Glass is a supercooled liquid.

(b) Markings on the glass plate are made by use of $$HF$$.

(d) Pyrex glass can withstand heat and contains $$B_2O_3$$.

(e) Opal glass contains $$CaF_2$$.

Helium can be excited to the $${1s}^{1}{2p}^{1}$$ configuration by light of $$58.44nm$$. The lowest excited singlet state, with the configuration $${ 1s }^{ 1 },{ 2s }^{ 1 }$$ lies $$4857{c}^{-1}$$ below $${1s}^{1}{2p}^{1}$$ state. What would be the average $$He-H$$ bond energy ($$kJ{ mol }^{ -1 }$$) have to be in order that $$He{H}_{2}$$ could form non-endothermically from $$He$$ and $${H}_{2}$$? Assume that the compound would form from the lowest excited singlet state of helium. Neglect any differences between $$\Delta U$$ and $$\Delta H$$. Take $$\Delta {H}_{f}(H)=218.0kj/mol$$.
[Give answer in nearest integer]

Formation of $$He{H}_{2}$$ requires energy equal to sum of
(i) Energy for excitation from $${1s}^{2}$$ to $${ 1s }^{ 1 },{ 2s }^{ 1 }$$ to form $$He$$ singlet is equal to: [Energy needed for excitation from $${1s}^{2}$$ to $${1s}^{1}{2p}^{1}$$-energy level difference in between $${1s}^{1}{2s}^{1}$$ and $${1s}^{1}{2p}^{1}$$]
Thus $${ E }_{ He }singlet=\cfrac { hc }{ { \lambda }_{ 1 } } -\cfrac { hc }{ { \lambda }_{ 2 } } =3.40\times { 10 }^{ -18 }-9.66\times { 10 }^{ -20 }J$$
$$=3.30\times { 10 }^{ -18 }J/molecule$$
where $${ \lambda }_{ 1 }=58.44\times { 10 }^{ -9 }m;\quad \cfrac { 1 }{ { \lambda }_{ 2 } } =4857{ cm }^{ -1 }$$
(ii) Energy to produce two mole of $$H$$,i.e $$2\times 218.0=436kJ/mol$$
Thus, $$E$$ for 2 mol bonds of $$He-H$$
$$=[3.30\times { 10 }^{ -18 }\times 6.023\times { 10 }^{ 23 }+436\times { 10 }^{ 3 }]J/mol$$
$$=2423.5kJ{mol}^{-1}$$
$$ { E }_{ He-H }=1211.8kJ{ mol }^{ -1 }$$

A certain salt (X) gives the following tests :
(i) Its aqueous solution is alkaline to litmus
(ii) On strongly heating it swells to give a glassy material.
(iii) When concentrated sulphuric acid is added to a hot concentrated solution of (X) , white crystals of a weak acid separate out
Write equations for all the above reactions and identify X, Y, Z.

The salt (X) is borax $$Na_2B_4O_7 .10H_2O$$
(i) In the aqueous solution it undergoes hydrolysis to form a mixture of boric acid and sodium hydroxide.
Hence, the aqueous solution is alkaline in nature.
$$Na_2B_4O_7 + 7H_2O \longrightarrow 4H_3BO_3 + 2NaOH \ or\ Na[B(OH)_4]$$
(ii) On strong heating, borax swels to a glassy material $$2NaBO_2 +B_2O_3$$
$$Na_2B_4O_7 \overset {\triangle}{\longrightarrow} \underbrace { 2NaBO_2 +B_2O_3 }_{ glassy } $$
(iii) When borax reacts with hot concentrated sulphuric acid solution, boric acid crystallizes out.
$$Na_2B_4O_7 . 10H_2O + H_2SO_4 \longrightarrow Na_2SO_4 + 4H_3BO_3 + 5H_2O$$

In some of the reactions thallium resembles aluminium, whereas in others it resembles with group I metals. Support this statement by giving some evidences.

Thallium is a group 13 element with common oxidation states +1 and +3.
Due to inert pair effect +1 oxidation state in Tl is more stable than +3 oxidation state. Al shows +3 oxidation state and alkali metals shows +1 oxidation state. Thus, thallium resembles both Al, and alkali metals.
Like Al, Tl forms compounds such as $$\displaystyle TlCl_3$$ and $$\displaystyle Tl_2O_3 $$ and like alkali metals, Tl forms compounds such as $$\displaystyle TlCl $$ and $$\displaystyle Tl_2O. $$

Explain the following reactions.
(a) Silicon is heated with methyl chloride at high temperature in the presence of copper.
(b) Silicon dioxide is treated with hydrogen fluoride.
(c) CO is heated with ZnO.
(d) Hydrated alumina is treated with aqueous NaOH solution.

(a) Silicon is heated with methyl chloride at high temperature in the presence of copper catalyst at 537 K, methyl substituted chlorosilanes $$\displaystyle MeSiCl_3, Me_2SiCl_2, Me_3SiCl$$ and $$Me_4Si $$ are formed.

(b) When silicon dioxide is treated with hydrogen fluoride, first $$\displaystyle SiF_4 $$ is formed and then hydro fluorosilicic acid is obtained.
$$\displaystyle SiO_2 + 4HF \rightarrow SiF_4 + 2H_2O $$
$$\displaystyle SiF_4 + 2HF \rightarrow H_2SiF_6 $$

(c) When CO is heated with ZnO, ZnO is reduced to Zn metal.
$$\displaystyle CO + ZnO \xrightarrow {\Delta} CO_2 + Zn$$

(d) When hydrated alumina is treated with aqueous NaOH solution, it dissolves to form sodium meta aluminate.
$$\displaystyle Al_2O_3.2H_2O + 2NaOH \rightarrow 2 NaAlO_2 + 3H_2O $$

A certain salt $$X$$, gives the following results.

(i) Its aqueous solution is alkaline to litmus.

(ii) It swells up to a glassy material $$Y$$ on strong heating.

(iii) When conc. $$H_2SO_4$$ is added to a hot solution of $$X$$, white crystal of an acid $$Z$$ separates out.

Write equations for all the above reactions and identify $$X, Y$$ and $$Z$$.

The compounds X, Y and Z are borax, sodium metaborate + boric anhydride and boric acid respectively.

When borax is heated, it first swells and then forms a transparent glass like bead of sodium meta borate and boric anhydride.

$$\displaystyle \underset {\text {Borax}}{Na_2B_4O_7} \xrightarrow {heat}
\underset {\text {sodium metaborate}}{2NaBO_2} + \underset {\text {Boric
anhydride}}{B_2O_3} + 10H_2O $$

Aqueous solution of borax is alkaline due to formation of strong base NaOH.

Hence, it turns red litmus blue.

$$\displaystyle Na_2B_4O_7 +7H_2O \rightarrow 4H_3BO_3 + 2NaOH $$

Borax reacts with sulphuric acid to form boric acid and sodium sulphate.

$$\displaystyle \underset {\text {Borax}}{Na_2B_4O_7} + H_2SO_4 +5H_2O \rightarrow \underset {\text {Boric acid}}{4H_3BO_3} + \underset {\text {Sodium sulphate}}{Na_2SO_4} $$

(a) Classify following oxides as neutral, acidic, basic or amphoteric:
$$CO,\, B_2O_3,\, SiO_2,\, CO_2,\, Al_2O_3,\, PbO_2,\, Tl_2O_3$$
(b) Write suitable chemical equations to show their nature.

The neutral oxide is CO.
The acidic oxides are $$\displaystyle B_2O_3, SiO_2$$ and $$ CO_2$$.
The basic oxide is $$\displaystyle Tl_2O_3 $$.
The amphoteric oxides are $$\displaystyle Al_2O_3 $$ and $$PbO_2 $$.

Number of $$B-O-B$$ bond in borax is ________.

From the structure, it is clear that Borax ($$Na_2B_4O_7$$) has 5 $$B-O-B$$ bonds.

Which is most abundant element in $$17th$$ group?

Fluorine exists as a diatomic molecule in its free state (F₂) and is the most abundant halogen found in the Earth's crust. Fluorine is the most electronegative element in the periodic table. It appears as a pale yellow gas at room temperature. Fluorine also has a relatively small atomic radius.

Explain the synthesis of ammonia occurs by Haber's process.

(a) Synthesis of ammonia occurs by Haberes process. Habers process is a reversible reaction between nitrogen and hydrogen to form ammonia using Iron as a catalyst.

$$3H_2(g) + N_2(g) \rightleftharpoons 2NH_3(g)$$ ; $$\triangle H= -92KJ$$

It is a exothermic reaction, as $$92KJ$$ of heat is given out. formation of ammonia through this process is followed by high pressure and low temperature

Explain: (i) $${[Si{F}_{6}]}^{2-}$$ is known whereas $${[Si{Cl}_{6}]}^{2-}$$ is not known.
(ii) Boron is unable to form $${[{BF}_{6}]}^{3-}$$ ion
(iii) conc. $$H{NO}_{3}$$ can be stored in Aluminium container.

a)Due to smaller size of F as compared to Cl , six small F atoms can be easily accommodated around Si atom while Cl cannot since F are present in smaller 2p orbital and Cl are present in larger 3p orbital therefore interaction of F lone pair electron with d-orbitals of silicon are stronger in F as compared Cl.

b)Boron does not have d orbitals. Because of this, it is not possible for the element to expand on its octet. This, in turn, causes the maximum covalence of boron to be restricted to 4. Beyond this value, boron will become unstable and not be able to form $$[BF_6]^{3-}$$ ions.

c)Nitric acid reacts with aluminium and forms a protective layer i.e. $$Al_2O_3$$. This protective layer makes aluminium unreactive towards $$HNO_3$$ and therefore, it can be stored in aluminium containers.

What happens when
i) White phosphorous is heated with concentrated $$NaOH$$ solution in an inert atmosphere of $$CO_2$$.
ii) $$XeF_6$$ undergoes partial hydrolysis.

A. 1.White phosphorous dissolves in boiling $$NaOH$$ solution in a atmospheric $$CO_2$$, to give

Phosphine, $$(PH_3)$$.

2.The equation is $$P_4 + 3NaOH + 3H_2O ------> PH_3 + 3NaH_2PO_2$$.

B. When xenon hexaflouride undergoes partial hydrolysis it giving

$$XeF_6 + H_2O → XeOF_4 + 2 HF$$

Which of the following reacts with water and aqueous solution become acidic $$SiCl_4$$ or $$CCl_{4}$$?

$$SiCl_4$$ reacts with water and aqueous solutions and it becomes acidic.
$$SiCl_4+2H_2O\longrightarrow SiO_2+4HCl$$
This reaction produce strong $$HCl$$, while $$CCl_4$$ does not react with water.

Write down the balanced equations for the reaction when calcium phosphate is heated with a mixture of sand and carbon.

The balanced equations for the reaction when calcium phosphate is heated with a mixture of sand and carbon is:

$$2Ca_3(PO_4)_2+6SiO_2+10C\rightarrow 6CaSiO_3+10CO+P_4$$

Calcium silicate, carbon monoxide and phosphorus are formed.

Name the metals with which the following ores/ minerals are associated :
Borax

We know that formula of borax is Na₂[B₄O₅(OH)₄]·8H₂O

So here Na ( Sodium) is a metal.

Match the list I and List II

1713507_1727093_ans_3ce5678434664fc08f945c0a2047848d.JPG

Explain why the tendency to show-$$2$$ oxidation state diminishes from sulphur to polonium?

On moving down the group, the size of the atom increases and the electronegativity decreases. Thus, the tendency to accept two more electrons as to attain octet in outermost shell decreases, i.e. the tendency to show $$-2$$ oxidation state decreases.

An inoganic halide $$(A)$$ reacts with water to form two acids $$(B)$$ and $$(C). (A)$$ also reacts with $$NaOH$$ to form two salts $$(D)$$ and $$(E)$$ which remain in solution. The solution gives white precipitates with both $$AgNO_{3}$$ and $$BaCl_{2}$$ solution respectively. $$(A)$$ is a useful organic reagent.

1661407_1727733_ans_061c6833345b408dab1a5d839dbdec83.jpg

Answer the following
Which element among group $$13$$ elements has the highest ionisation enthalpy?

On moving down the group from

B

T l

, a regular decreasing trend in the I.E. is not observed.

Element	$B$	$A l$	$G a$	$I n$	$T l$
I.E. (kJ/mol)	801	577	579	558	589

In Ga, there are ten d-electrons in the penultimate shell which screen the nuclear charge less effectively and thus, outer electron is held firmly. As a result, the ionisation energy of both Al and Ga is nearly the same. The increase in ionisation energy from In to Tl is due to poor screening effect of 4f electrons present in the inner shell.

Thus, the correct order is,

B > A l < G a > I n < T l

Hence, correct answer is Boron.

Why $$BiH_3$$ is the strongest reducing agent?

The reducing character of the hydrides of Group 15 elements increases from NH₃ to BiH₃(Bismuthine) because the reducing character depends upon the stability of the hydride.The greater the unstability of hydride,the greater is the its reducing character. Since,the BiH₃ is the least stable (because the size of central atom is greatest & therefore its tendency to form stable covalent bond with small hydrogen atom decreases,as a result the bond strength decreases) in this series,hence the reducing character increases.

Give reason for each of the following:
$$SnCl_2$$ is a solid while $$SnCl_4$$ is a liquid at room temperature.

The size of $$ Sn^{2+}$$ ion is larger than $$Sn^{4+}$$ ion, so $$SnCl_2$$ is an ionic compound and $$SnCl_4$$ is a covalent compound. Thus $$SnCl_4$$ molecule are held together by weak van der Waals forces. Hence $$SnCl_4$$ exist as a liquid.

On the other hand, $$SnCl_4$$ molecule are held with the help of strong electronic forces between the opposite charges ion. So $$SnCl_2$$ is solid.

Starting from borax, how would you obtain?
Boron trioxide

1620051_1728062_ans_f7fa683420ec43f0871a7fe26d5e2cfc.jpg

Answer the following
Which has higher ionisation enthalpy?
$$Al \ or \ Ga$$

Answer:Gallium

Explanation:

The trend in ionization enthalpy is because there is decrease in Ionisation enthalpy from B to Al due to increase in size and from Al to Ga. Ga has more ionisation energy than Al due to uneffective screening effect.

Answer the following
What is Tincal?

It is a mineral of boron. Tincal is a crude type of borax, a cleaning material.

Its composition is $${Na}_{2}{B}_{4}{O}_{7}.{10H}_{2}O$$.

Explain the following with relevant reason.
$${AlCl}_{3}$$ forms a dimer but $${BCl}_{3}$$ does not form dimer.

$${AlCl}_{3}$$ lacks back bonding as in $${BCl}_{3}$$ because of increase in size of aluminium. Aluminium metal atoms complete their octet by coordinate bond forming bridges by chlorine atoms between two $$Al$$ atoms.
1623644_1729334_ans_2c6d4082a6624e8a95bb905d2b9183df.jpg

1623644_1729334_ans_2c6d4082a6624e8a95bb905d2b9183df.jpg

Identify the unknown species and complete the following equation:

$$Na_2B_4O_10H_2O+ conc. H_2SO_4 \xrightarrow [ ]{ Heat } (A)$$

Accordomg to the question,

Borax on heating with con. sulphuric acid gives a compound (A). The whole reaction can be written as:

$$Na_{2}B_{4}O_{7}.10H_{2}O + conc.H_{2}SO_{4} \overset{Heat} \rightarrow Na_{2}SO_{4} +4H_{3}BO_{3} + 5H_{2}O$$

So, $$(A) = H_{3}BO_{3}$$

Describe the preparation, properties, and uses of the following compound
Borax.

1670662_1729958_ans_608c10951aea4c659b361051ef417d4a.jpg

A colourless solid $$(A)$$ on hydrolysis produces a heavy white precipitate $$(B)$$. Solid $$(A)$$ gives a clear solution in conc. $$HCl$$; however when added to large amount of water, it again gives precipitate of $$(B)$$. When $$H_2S$$ passed through a suspension of $$(A)$$ or $$(B)$$, a brown black precipitate of $$(C)$$ is obtained. Compound $$(A)$$ liberates a gas $$(D)$$ on treating with $$H_2SO_4$$. The gas $$(D)$$ is water soluble and gives white precipitate $$(E)$$ with solution of mercurous salt but not with mercuric salt. Identify $$(A)$$ to $$(E)$$.

$$(A) BiCl_2; (B) BiOCl; (C) Bi_1S_3; (D) HCl; (E) Hg_2Cl_2$$

When aqueous solution of borax is acidifies with hydrochloric acid, a white crystalline solid is formed which is soapy to touch. Is this solid acidic or basic in nature? Explain

1895419_1729581_ans_172fe80f0a904379986f17253e6dd148.jpg

The +1 oxidation state in group 13 and +2 oxidation state in group 14 becomes more and more stable with increasing atomic number. Explain.

In group 13 and 14 , as we move down the group, the tendency of $$8-e^{-}$$ of valence shell to participate in bond formation is due to ineffective shielding of s -e's of valence shells by intervening of d & f electrons. This is called inert pair effect.

Due to this, s-electrons of valence shell of group 13 and 14 are unable to participate in bonding. Hence, +1 and +2 oxidation states, in group 13 and 14 respectively becomes more stable with increasing atomic number.

Explain the following with relevant reason.
Why boron does not form $${B}^{3+}$$ ion?

The total sum of all the three ionisation energies of boron is so high that lattice energy liberated by interaction of $${B}^{3+}$$ and anions is not enough for formation of this ion.

Give formula of the following:

Rasorite

1670667_1729942_ans_e0952375af7946b4a7e5379c6bed0131.jpg

Properties of borax are represented by the following road map. Identify the compounds $$'A'$$ to $$'D'$$ and explain the reaction involved.

1731802_1733726_ans_010e3584bfe445609173f104648af54a.jpg

What happens when:
A mixture of borax and cobalt oxide is heated in a flame.

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Show the effect of heat on the following commercially important compounds:

Borax, $$Na_2B_4O_10H_2O$$.

When heated, borax undergoes various transitions. It first loses water molecules and swells. Then, it turns into a transparent liquid, solidifying to form a glass-like material called borax bead.

$$\underset{Borax}{Na_2B_4O_7.10H_2O}\rightarrow \underset{Sodium\,\,Metaborate}{Na_2B_4O_7}\rightarrow 2NaBO_2+\underset{Boric\,anhydride}{B_2O_3}$$

A white crystalline solid $$'A'$$ dissolves in water to give an alkaline solution. On heating $$'A'$$ first loses water molecules and swells up. On further heating it turns into a transparent liquid which solidifies into a glassy bead.

$$'A'$$ is Borax i. e. $$Na_2[B4O_5(OH)_4].8H_2O$$ which is also written as.

The reactions are:

$$Na_2B_4O_7 + 7H_2O \rightarrow 2NaOH + 4H_3BO_3$$ Orthoboric acid

On heating:

$$Na_2B_4O_7.10H_2O \overset{\Delta }{\rightarrow} Na_2B_4O_7 \overset{\Delta }{\rightarrow} 2NaBO_2 + B_2O_3$$

Borax Sodium metaborate

Transparent glassy bead

Explain the chemistry of borax bead test.

Borax Bead Test: It is a chemical test to disclose the presence of certain metals in a sample. A simple laboratory test for certain metal ions in salts. A small amount of the salt is mixed with borax and a molten bead formed on the end of a piece of platinum wire. Certain metals can be identified by the colour of the bead produced in the oxidizing and reducing parts of a Bunsen flame.

Sodium tetaraborate decahydrate $$Na_2B_4O_7.10H_2O\rightarrow Na_2B_4O_7\rightarrow 2NaBO_2+B_2O_3$$

$$B_2O_3$$ combines with metal oxides to form metal metaborates which form coloured beads. This reaction is called "Borax bead test"

$$B_2O_3+CoO\rightarrow Co(BO_2)_2$$

In the following reactions assign for underlined atom for product of complete hydrolysis at R.T.
A. If product is oxy acid with —ic suffix.
B. If product is oxy acid with —ous suffix.
C. If product are two oxy acids one with —ic suffix and another one with —ous suffix.
D. If product is not oxy acid, neither with —ic suffix nor with —ous suffix.
$$H_4\underline B_2O_5+H_2O \longrightarrow H_3BO_3$$

1937266_1862057_ans_7211ae57925940b0bd34d18a21040a30.jpg

Symbol	.................
Atomic number	.................
No. of electrons in the outermost shell	.................
Valency	.................
Metal/ non-metal?	.................

The P-Block Elements - Class 11 Medical Chemistry - Extra Questions