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CBSE Questions for Class 12 Engineering Chemistry Coordination Compounds Quiz 5 - MCQExams.com
CBSE
Class 12 Engineering Chemistry
Coordination Compounds
Quiz 5
The synthetic steroid ethynylestardiol (1) is a compuond used in the birth control pill. How many $$sp^2$$ hybridised carbon atoms are present in compound (1)?
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0%
4
0%
5
0%
6
0%
7
0%
8
Explanation
$$sp^2$$ hybridised carbon are those carbon which form double bond
(Refer to Image)
There re $$6$$ $$sp^2$$ hybridised carbon.
The primary and secondary valency of a central metal ion in $$[Co(NH_3)_4CO_3]$$$$Cl$$ complex is _________ and _________ respectively.
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$$2, 6$$
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$$4, 6$$
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$$4, 5$$
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$$3, 6$$
Explanation
The primary and secondary valency of cobalt ion is 3 and 6 respectively.
The charge on cobalt ion is $$+3$$ which balances $$-2$$ charge on carbonate ion and $$-1$$ charge on chloride ion. Hence, the primary valency of cobalt ion is 3.
Cobalt ion is surrounded by four monodentate and one bidentate ligand (4 ammonia and on carbonate) in the coordination sphere. Hence, its coordination number (secondary valency) is 6.
By which of the following, poisoning of lead in the body can be removed?
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$$Pt$$
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EDTA
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$$Ox^{2-}$$
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$$Pn$$
Explanation
EDTA is a hexadentate ligand and forms the water-soluble stable chelate complex with metals such as lead. Hence, it can be used to remove lead from the body. In this way, EDTA helps in removing poisoning of lead from the body.
Which of following is the incorrect match?
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Insulin - Zinc
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Haemoglobin - Iron
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Vitamin $$B_{12}$$ - Cobalt
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Chlorophyll - Chromium
Explanation
Chlorophyll contains magnesium ion, encased in a large ring structure known as chlorine, derived from pyrrole.
Order of the bond strength of $$C-H$$ bonds involving sp, $$sp^2$$ and $$sp^3$$ hybridized carbon atoms is:
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$$sp \, > \, sp^2 \, > \, sp^3$$
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$$sp^3 \, > \, sp^2 \, > \, sp$$
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$$sp^2 \, > \, sp^3 \, > \, sp$$
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$$sp^2 \, > \, sp \, > \, sp^3$$
Explanation
$$\equiv \overset {sp}{C}-H$$ $$=\overset {sp^2}{CH}-H$$ $$-\overset {sp^3}{CH_2}-H$$
In $$sp, sp^2$$ and $$sp^3$$ hybridized central carbon atoms, the percentage of $$s$$ character in bonding orbitals (i.e. $$C-H$$ bonds) are $$50$$%, $$33.3$$% and $$25$$% respectively. We know that percentage of $$s$$ character increases the bond strength because the shape of $$s$$ orbitals is spherical and it is more closer to nucleus of an atom. Therefore order of bond strength of $$C-H$$ bonds will be:
$$sp >sp^2 > sp^3$$
The synthetic steroid ethynylestardiol (1) is a compuond used in the birth control pill.
How many $$sp$$ hybridised carbon atoms are present in compound (1) ?
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2
0%
4
0%
6
0%
8
0%
10
Explanation
$$sp$$ hybridized carbon are those who forms triple bond.
(Refer to Image)
These are two carbons which are $$sp$$ hybridized.
Which of the following will exhibit more than ten isomers?
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$$\left[ Co\left( en \right) Cl\ BrI\left( CN \right) \right]$$
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$$\left[ Co\left( gly \right) Cl\ BrI\left( NC \right)\right]$$
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$$\left[ Co\left( gly \right) Cl\ Br\left( CN \right)NO_2 \right]$$
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All the above
Explanation
There are five types of ligand present in which one is bidentate and one is ambidentate i.e. $$(CN)$$.
(Refer to Image)
These are $$6$$ basic isomers. As one ambidentate ligand is present i.e. $$CN$$ or $$NO_2$$ which can attach through two site. Thus show linkage isomerism.
Thus the total number of isomers will be greater than $$10$$. Thus all will give more than $$10$$ isomers.
Which of the following does not show optical isomerism?
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$$[Co(NH_3)_3Cl_3]$$
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$$[Co(en)Cl_2(NH_3)_2]^+$$
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$$[Co(en)_3]^{3+}$$
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$$[Co(en)_2Cl_2]^+$$
Explanation
The compound $$[Co(NH_3)_3Cl_3]$$ does not show optical isomerism
.
This is a coordination compound where coordination number is 6 and geometry octahedral symmetrical. Compounds with an octahedral geometry having bidentate ligands show optical isomerism.
The given compound does not contain any bidentate ligands, also it has symmetrical geometry, thus, it does not show optical isomerism.
The value for crystal field stabilisation energy is zero for:
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$$K_2(MnF_6]$$
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$$K_3[Fe(CN)_6]$$
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$$K_3[FeF_6]$$
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$$K_4[Fe(CN)_6]$$
Explanation
Solution:- (C) $${K}_{3} \left[ Fe{F}_{6} \right]$$
The CFSE for tetrahedral complex is given by-
$$CFSE = -0.4 {t}_{{2}_{g}} {e}^{-} + 0.6 {e}_{g} {e}^{-}$$
For $${K}_{3} \left[ Fe{F}_{6} \right]$$-
$$\Rightarrow {Fe}^{+3} = {{t}_{{2}_{g}}}^{3} {{e}_{g}}^{2}$$
$$CFSE = -0.4 \times 3 + 0.6 \times 2 = 0$$
Which of the following order is correct for the IR vibrational frequency of CO?
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$${ \left[ Fe{ \left( CO \right) }_{ 4 } \right] }^{ 2- }<\left[ Co{ \left( CO \right) }_{ 4 } \right]^- <\left[ Ni{ \left( CO \right) }_{ 4 } \right] $$
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$${ \left[ Fe{ \left( CO \right) }_{ 4 } \right] }^{ 2- }>\left[ Co{ \left( CO \right) }_{ 4 } \right]^- >\left[ Ni{ \left( CO \right) }_{ 4 } \right] $$
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$${ \left[ Fe{ \left( CO \right) }_{ 4 } \right] }^{ 2- }>\left[ Co{ \left( CO \right) }_{ 4 } \right]^- <\left[ Ni{ \left( CO \right) }_{ 4 } \right] $$
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$${ \left[ Fe{ \left( CO \right) }_{ 4 } \right] }^{ 2- }<\left[ Co{ \left( CO \right) }_{ 4 } \right]^- >\left[ Ni{ \left( CO \right) }_{ 4 } \right] $$
Explanation
The IR vibrational frequency depends on the strength of CO bond, More is the bond strength more will be the IR vibrational frequency.Since, CO is a pi acceptor ligand and accepts electron in antibonding pi orbital. More electrons in antibonding orbital implies less bond strength of CO. In the given complexes the oxidation state of metals is Fe(2-), Co(1-), Ni(0). Therefore more electrons on metal implies more will be the pi backbonding more stronger will be the Metal-C bond weaker will be the C-O bond hence, the IR vibrational frequency order will be $$[Fe(CO)_4]^{2-} < [Co(CO)_4]^{-} < [Ni CO)_4]$$
The correct representation of $$CuSO_4\cdot 5H_2O$$ is:
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$$[Cu(H_2O)_5]SO_4$$
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$$[Cu(H_2O)_3SO_4]\cdot 2H_2O$$
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$$[CuSO_4\cdot H_2O]\cdot 4H_2O$$
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$$[Cu(H_2O)_4]SO_4\cdot H_2O$$
Explanation
For,
$$CuS{O_4}.5{H_2}O$$ in this $$4{H_2}O$$ are bonded by co-ordinate bond and $$1 {H_2}O$$ by covalent bonding.
Therefore,
Its formula of the compound is $$\left[ {Cu{{\left( {{H_2}O} \right)}_4}} \right]S{O_4}.{H_2}O$$.
Which of the following alkaline earth metal sulphates has hydration enthalpy higher than the lattice enthalpy ?
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$$SrSO_4$$
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$$MgSO_4$$
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$$CaSO_4$$
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None of these
The value of X in $$Fe{ \left( CO \right) }_{ 2 }\left( NO \right) x,$$ is:
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0%
3
0%
4
0%
2
0%
1
Explanation
The coordination sphere of a coordination compound or complex consists of the central metal atom/ion plus its attached ligands. The coordination sphere is usually enclosed in brackets when written in a formula. The coordination number is the number of donor atoms bonded to the central metal atom/ion.
Most metal complexes or compounds except for alloys. Specific examples include hemoglobin and $$Ru_3(CO)_{12}$$.
So the total electron should be 18
valence electron of iron is = 8
CO gives = $$2e^-$$ , NO gives $$3e^-$$
18 = $$8+(2\times2)+(3\times x)$$
18 - 8 - 4 = $$3x$$
$$x$$ = 2
$$NiCl^+_2\overset{KCN}{\underset{excess}{\rightarrow}}Complex(A)$$
$$NiCl^+_2\overset{KCl}{\underset{excess}{\rightarrow}}Complex (B)$$
The coordination number of nickel in both complexes (A) and (B) is $$4$$.
The IUPAC name of the complex (A) and (B) are:
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Potassium tetracyanonicklate (II) and Potassium tetrachloronicklate (II)
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Potassium tetracyanonickel (II) and Potassium tetrachloronickel (II)
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Potassium cyanonicklate (II) and Potassium chloronicklate (II)
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Potassium cyanonickel (II) and Potassium Chloronickel (II)
Which of the following compounds is most stable ?
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$$LiCl$$
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$$LiI$$
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$$LiBr$$
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$$LiF$$
Explanation
$$\bf{Explanation-}$$
Lithium is an alkali metal and it is electro-positive in nature.
The order of electronegativity of halides present in these compounds are as follow:
$$F>Cl>Br>I$$
Hence, lithium fluoride $$(LiF)$$ is the most stable compound b
ecause of the small size of $$Li$$.
As we go down the group in the periodic table the size of the element increases so, the $$Li-X$$ bond becomes weak.
As $$Li$$ and $$F$$ are of comparable size that is why $$LiF$$ is the most stable.
Also, $$LiF$$ has the highest lattice enthalpy.
$$\bf{Conclusion-}$$ Hence, the correct answer is option D.
Green house effect is caused by
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$$NO_2$$
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$$CO_2$$
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$$NO$$
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All are correct
Which of the following can show coordination isomerism ?
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$$[Cu(NH_3)_4][PtCl_4]$$
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$$[Fe(NH_3)_6]^{-2}[Pt(CN)_6]^{-3}$$
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Both A and B
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$$[Pt[en]_3] [SO_4]_2$$
Consider the reaction.
$$[Ni(H_2O)_6]^{2+}(aq)+3en(aq) \to [Ni(en)_3]^{2+}+2H_2O$$
Which of the following property of compound is changed in reaction?
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Co-ordination number
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Geometry of complex
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Hybridization of central atom
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Wavelength of absorbed light
Explanation
In the given reaction the only property that is changing is the wavelength of absorbed light because at the reactant side the complex have water as a ligand which is a weak field ligand on the other side the the product formed has the ethylenediammine ligand attached to metal which is a strong field ligand. As the strength of the ligand changes the $$\Delta_o$$ value changes and hence the wavelegth which is indirectly proportional to it also changes. Hence D is the correct option
If the thio-cyanide ion is added to potash-ferric alum then red colour appears. This colour is due to the formation of:
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$$KSCN$$
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$$Fe(SCN)_3$$
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$$Fe(SCN)_2$$
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$$Fe(SCN)$$
Explanation
The $$Fe^{3+}$$ ions react with the thiocyanide ion present in the solution to form a dark red colored complex of iron thiocyanate, i.e;
$$Fe^{3+}+SCN^-+5H_2O\longrightarrow [Fe(SCN)$$3$$(H_2O)_5]^{2+}$$
Number of stereoisomer's possible for coordination complex $$Na[Cr(en)Cl_2Br_2]$$ is:
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2
0%
3
0%
4
0%
5
Explanation
There are four stereoisomers are possible for the given coordination compound.
It shows optical as well as geometrical isomerism.
The co-ordination number of a metal in co-ordination compound is:
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same as primary valency
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sum of primary and secondary valences
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same as secondary valency
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none of the above
Explanation
The secondary valency
is equal to the coordination number
the
secondary valency
are non ionizable
valencies. These are satisfied by neutral molecules or negative ions. For example in $$[Ni(CO)_4]$$ the coordination number of Ni metal is four and its secondary valency is also four.
An ion $${ M }^{ 2+ }$$, forms the complexes $${ \left[ M{ \left( { H }_{ 2 }O \right) }_{ 6 } \right] }^{ 2+ },{ \left[ M{ \left( en \right) }_{ 3 } \right] }^{ 2+ }$$ and $${ \left[ { MBr }_{ 6 } \right] }^{ 4- }$$. The colour of the complexes will be _____________ respectively.
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green, blue and red
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blue, red and green
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green, red and blue
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red, blue and green
Explanation
Crystal Field Stabilization Energy(CFSE) is proportional to the frequency of the absorbed light. The emitted colors are red, green and blue. The corresponding absorbed colors are green, red and orange. So the first complex will be blue, second is red and third is green.
The coordination number and oxidation number of the central metal ion in the complex $$[Pt(en)_2]^{+2}$$ is:
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C.N. $$=2$$, O.N. $$=+2$$
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C.N. $$=6$$, O.N. $$=+4$$
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C.N. $$=4$$, O.N. $$=+4$$
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C.N. $$=4$$, O.N. $$=+2$$
Explanation
Ethylenediamine which is bidentate ligand is of two in number hence 2X2 =4
Hence co-ordination number of Platinum will be 4
Oxidation state = [x + 2 (0)] = 2+
x =+2
Thus, coordination number is 4 and oxidation number is +2
How many of the following metals when heated in an atmosphere of $$N_2$$ gas form nitrides?
Li, Na, K, Rb, Cs, Mg, Ca, Sr, Ba
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9
0%
5
0%
3
0%
6
$$FeCl_4H_2O$$ is actually:
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$$[Fe(H_2O)_4]Cl_3$$
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$$[Fe(H_2O)_3Cl]Cl_2.H_2O$$
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$$[Fe(H_2O)_4Cl_2]Cl$$
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$$[Fe(H_2O)_3Cl_2]Cl.H_2O$$
Explanation
$$Fe{Cl}_{3}$$.$$4{H}_{2}O$$ is actually $$[Fe({H}_{2}O)_{4}]{Cl}_{3}$$.
The water molecules are coordinated with the iron atom, are inside the coordination sphere whereas the chloride ions are outside the Ionization sphere.
Hence, the correct option is A
Which of the following is $$correct\ IUPAC$$ name for coordination isomer of $$[Pt(NH_{3})_{4}][Pd(C_{2}O_{4})_{2}]$$?
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Diammine (oxalato) platinum (II) diammine (oxalato) paladate (II)
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Diammine (oxalato) platinum (II) diammine (oxalato) paladium (II)
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Tetraammineplatinum (IV) bis(oxalato) paladate (II)
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Tetraammineplatinum (IV) bis(oxalato) paladium (II)
Explanation
Rules for writing IUPAC name in Coordination Chemistry
:-
Rule 1- While naming a coordination compound cation is always named before the anion irrespective of the fact that whether complexion is cation or anion.
Rule 2- If there are more than one type of ligands present in any coordination compound, the name of the ligands is to be done in alphabetic order followed by the name of central metal atom/ion.
Name of anionic ligands ends with $$O$$.
For neutral ligands, the common name is used.
Rule 3- If the name of ligand already contains a numerical prefix, then the terms, bis, tris, tetrakis are used and the ligand to which they refer to being placed in parentheses.
Rule 4- After naming the ligand in alphabetical order name of the central metal atom/ion is written.
If the complex is an anion, the name of metal ends with the suffix -ate for Latin name.
Rule 5- The oxidation state of the metal in the complex is given as a Roman numeral in parentheses.
Rule 6- The neutral complex molecule is named similar to that of the complex cation.
Rule 7- These are some ligands that may be attached to the central metal atom/ion through different atoms. For eg- $$NO_2,CN,$$etc.
Like,
$$M-NO\longrightarrow$$ nitro
$$M-ONO^2\longrightarrow$$ nitrito
$$M-SCN\longrightarrow$$ thiocyanato
$$M-NCS\longrightarrow$$ isothiocyanato
IUPAC name of $$[Pt(NH_3)_4][Pd(C_2O_4)_2]$$
Tetraamine platinum (IV) bis-oxalatopaladate (II)
Primary and secondary valency of $$Pt\ in\ \left[ Pt{ \left( en \right) }_{ 2 }{ Cl }_{ 2 } \right] { Cl}_{ 2 }$$ are:
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4, 4
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4, 6
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6, 6
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4, 2
Explanation
Primary valency is the number of ions required to satisfy the charge on the metal ion.
Secondary valency is the number of ions of molecules that are coordinated to the metal ion.
So, in $$[Pt(en)_2Cl_2]Cl_2$$
Charge on $$Pt\Rightarrow x+2(0)+2(-1)=+2\Rightarrow x=+4$$
So, Primary valency is $$4$$
Secondary valency is $$2(2)+2=6$$
$$\therefore$$ Answer is option B.
Which of the following complex ion is expected to absorb light in the visible region?
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$$[Ti(en)_{3}]^{4+}$$
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$$[Sc(NH_{3})_{4}(H_{2}O)_{2}]^{3+}$$
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$$[CrO_{4}]^{2-}$$
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$$[Zn(CN)_{2}]^{2-}$$
Explanation
Only those transition metal complexes are expected to absorb visible light, in which d- subshell is incomplete and excitation of an electron from a lower energy orbital to higher energy orbital is possible.
(a) In $$\left[ Sc({ H }_{ 2 }O)({ NH }_{ 3 })_{ 3 } \right] ^{ 3+ }$$ Sc is present in $$ { Sc }^{ 3+ }$$=$$[Ar]3{ d }^{ 0 },4{ s }^{ 0 }$$
Since in this complex excitation of the electron is not possible, it will not absorb visible light.
(b) $$\left[ Ti(en)_{ 2 }(N{ H }_{ 3 })_{ 2 } \right] ^{ 4+ },$$ Ti is present as $${ Ti }^{ 4+ }$$
$${ Ti }^{ 4+ }=[Ar]3{ d }^{ 0 },4{ s }^{ 0 }$$
Hence it will not absorb visible light
(c) In $$\left[ Cr({ NH }_{ 3 })_{ 6 } \right] ^{ 3+ }$$ Cr is present as $${ Cr }^{ 3+ }$$
$${ Cr }^{ 3+ }$$=$$[Ar]3{ d }^{ 3 },4{ s }^{ 0 }$$
Since this complex has three unpaired electrons excitation of electrons is possible and thus, it is expected that this complex will absorb visible light.
(d) In $$\left[ Zn(N{ H }_{ 3 })_{ 6 } \right] ^{ 2+ }$$ Zn is present as $${ Zn }^{ 2+ }$$
$${ Zn }^{ 2+ }$$=$$[Ar]3{ d }^{ 10 },4{ s }^{ 0 }$$
Hence , this complex will not absorb visible light.
The correct option is C.
In the compound $$CoCl_{3}.5NH_{3}$$:
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all the $$Cl$$ show primary valency only
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Two $$Cl$$ show primary valency and one $$Cl$$ shows secondary valency
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Two $$Cl$$ show primary valency and one $$Cl$$ shows primary valency as well as secondary valency
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All the $$Cl$$ show secondary valency
Explanation
In the given coordination complex 5 $$NH_3$$ molecule are directly attached to metal whereas only one one Cl is directly attached to metal and rest 2 are satisfying the oxidation state of metal and called as primary valency of the metal.Therefore it can be seen from the image and the formula of the complex which is $$[Co(NH_3)_5Cl]Cl_2$$ that the one Cl is acting as a secondary valency and 2 Cl act as primary valency.Hence option B is correct.
Which of the following will given maximum number of isomers?
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$$[Co(NH_3)_4 Cl_2]^+$$
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$$[Ni(en)(NH_3)_4]^{2+}$$
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$$[Cr(SCN)_2(NH_3)^+_4]$$
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$$[Ni(C_2O_4)(en)_2]^{2+}$$
Explanation
It can cis-trans isomers and also has stereo isomers in the trans state while others have only two isomeric structures.
1 mole of co-ordinate compound $$CoCl_3$$ on reaction with excess AgNCl $$143.5$$gms of AgCl ppt . Then number of chloride ions satisfying both primary valency and secondary valency are?
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$$2$$
0%
$$3$$
0%
$$1$$
0%
$$0$$
Explanation
Moles of $$AgCl$$ formed $$=\cfrac{143.5}{143.5}=1$$
$$1$$ mole of $$CoCl_3$$ gives $$1$$ mol of $$AgCl$$ ppt.
Secondary valence is the number of ions that are coordinated to metal ion. Here, $$1Cl^-$$ ion is not coordinated to the metal ion, due to which it gives $$AgCl$$ ppt. Hence, no. of chloride ions satisfying the given condition $$=2.$$
If a complex $$[M(H_2O)_6]^{2+}$$ absorbs energy corresponding to $$\lambda$$ nm, then the wavelength (in nm) absorbed by tetrahedral, complex $$[m(H_2O)_4]^{2+}$$ will be:
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$$\frac{4}{9}\lambda$$
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$$\frac{5}{9}\lambda$$
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$$\frac{9}{4}\lambda$$
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$$\frac{1}{\lambda}$$
Explanation
In coordination complex Energy is directly related to $$\Delta$$ and inversely to $$ \lambda $$. and it is known that
$$\Delta_t$$ =(4/9)
$$\Delta_o$$ then,
$$1/\lambda_t$$ =(4/9)
$$\frac{1}{\lambda _o}$$ Hence,
$$\lambda _t = \frac{9}{4}\lambda _o$$. this means The wavelegth absorbed by tetrahedral comlex will be 9/4
$$ \lambda $$.
Which of the following represents the given mode of hybridization $$sp^2 - sp^2 - sp - sp$$ from left to right?
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$$CH_2 = CH - C \equiv N$$
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$$CH_2 = C = C = CH_2$$
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$$CH_2 = CH - CH = CH_2$$
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$$CH_2 = C = CH - CH_3$$
Explanation
The type of hybridization of carbon atom is :
(A) $$\overset{{sp^2}}{CH_2}$$=
$$\overset{sp^2}{CH}$$-
$$\overset{sp}{CH}$$ = $$\overset{sp}{N}$$
(B)
$$\overset{sp^2}{CH_2}$$ =
$$\overset{sp}{CH}$$=
$$\overset{sp}{CH}$$ =
$$\overset{sp^2}{CH_2}$$
(C)
$$\overset{sp^2}{CH_2}$$=
$$\overset{sp^2}{CH}$$-
$$\overset{sp^2}{CH}$$=
$$\overset{sp^2}{CH_2}$$
(D)
$$\overset{sp^2}{CH_2}$$=
$$\overset{sp}{CH}$$=
$$\overset{sp^2}{CH}$$-$$\overset{sp^3}{CH_3}$$
Therefore, (A) has the given mode of hybridization from left to right.
Predict the order of $$\Delta_0$$ for the following compounds.
I. $$[Mn(H_2O)_6]^{2+}$$
II. $$[Mn(CN)_2(H_2O)_4]$$
III. $$[Mn(CN)_4(H_2O)_2]^{2-}$$
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$$\Delta_0 (I) < \Delta_0 (II) < \Delta_0 (III)$$
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$$\Delta_0 (II) < \Delta_0 (I) < \Delta_0 (III)$$
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$$\Delta_0 (III) < \Delta_0 (I) < \Delta_0 (I)$$
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$$\Delta_0 (I) < \Delta_0 (III) < \Delta_0 (II)$$
Explanation
More is the number of strong field ligand higher is the crystal field stabilisation energy.
So, $$[Mn(Cn)_4(H_2O)_2]^{2-}$$ having $$4$$ $$CN^-$$ ligands will have the highest value of $$\triangle_0$$ then followed by $$[Mn(CN)_2(H_2O)_4]$$ then $$[Mn(H_2O)_6]^{2+}$$.
The CFSE of $$[Mn(H_2O)_6]^{2+}$$ is:
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-1.2 $$\Delta_{\Theta}$$
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-1.9 $$\Delta_{\Theta}$$
0%
0
0%
-2.4 $$\Delta_{\Theta}$$
Explanation
The CFSE of a $$d^5$$octahedral complex is = -0.4$$\Delta_o$$x number of electrons in $$t_2g$$ orbitals + 0.6
$$\Delta_o$$
x number of electrons in $$eg$$ orbitals
=
-0.4$$\Delta_o$$x3 +
0.6
$$\Delta_o$$
x 2 = 0. hence the correct option is C.
The formula of the complex, tris- (ethylenediamine) cobalt (iii) sulphate is:
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$$ [Co(en)_{2}SO_{4}] $$
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$$ [Co(en)_{3}SO_{4}] $$
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$$ [Co(en)_{3}]_{2}(SO_{4})_{3} $$
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$$ [Co(en)_{3}]_{2}SO_{4} $$
Explanation
$${ \left[ CO{ \left( en \right) }_{ 3 } \right] }_{ 2 }{ \left( { SO }_{ 4 } \right) }_{ 3 }$$
$$\rightarrow$$ Oxidation of $$en=0$$ $${ { SO } }_{ 4 }=-2$$ then we get $$CO=+3$$ as given
$$\rightarrow$$ And tris is used as two numbers are present at that same spot.
$$\rightarrow$$ where $$2$$ is called $$"di"$$ and $$3$$ is called $$"tris"$$.
Ans :- Option C.
The pair(s) of coordination complex/ion exhibiting the same kind of isomerism is (are)?
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$$ [Cr(NH_3)_5Cl]Cl_2\ and\ [Cr(NH_3 )_4Cl_2]Cl$$
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$$[Co(NH_3)_4Cl_2]^+ and\ [Pt(NH_3)_2(H_2O)Cl]^+$$
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$$[CoBr_2Cl_2]^{2-} and\ [PtBr_2Cl_2]^{2-}$$
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$$ [Pt(NH_3)_3(NO_3)]Cl\ and\ [Pt(NH_3 )_3Cl]Br$$
The number of isomers possible for square planar complex $$ K_{2}[PdClBr_{2}SCN] $$ is:
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0%
2
0%
3
0%
4
0%
6
Explanation
The given planar complex can have upto two optical isomes 1. Cis-$$[PdClBr_2SCN]^{2-}$$ and 2.trans-
$$[PdClBr_2SCN]^{2-}$$.In first complex the Br-Br will be cis to each other and in second isomer they are trans to one another. Hence the correct option is A.
The strongest -CO bond is present in:
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$$ [Cr(CO)_{6}]^{+}$$
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$$ [Fe(CO)_{5}]$$
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$$ [V(CO)_{6}]^{-}$$
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all have equal strength
Explanation
The strongest CO bond is present in the complex$$ [Cr(CO)]^+$$. This is because as the oxidation state of metal increases the pi back bonding tendency becomes low and hence the strength of CO bond increases. In the given complexes the oxidation state increase in of the order $$V(-)<Fe(0)<Cr(+)$$. hence the correct option is A.
Which of the following complexes has a geometry different from others?
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$$\left [ NiCl_{4} \right ]^{2-}$$
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$$Ni\left ( CO \right )_{4}$$
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$$\left [ Ni\left ( CN \right )_{4} \right ]^{^{-2}}$$
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$$\left [ Zn\left ( NH_{3} \right )_{4} \right ]^{2+}$$
The number of isomers represented by the M.F $${C}_{2}{D}_{2}(Cl)(Br)$$ is:
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2
0%
3
0%
5
0%
none of these
Explanation
There are $$2$$ isomers possible.
Type of isomerism shown by $$\left [ Cr(NH_{3})_{5}NO_{2} \right ]Cl_{2}$$ is
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Optical
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ionisation
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geometrical
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linkage
Explanation
This complex has $$NO_2$$ in it,which is an ambidentate ligand,which means it has $$2$$ sites of attack to the central metal atom.
Therefore,the complex which has such type of ligand can show linkage isomerism and they are
(1)$$[Cr(NH_3)_5NO_2]Cl_2$$
(2)$$[Cr(NH_3)_5ONO]Cl_3$$
oxidation state of chromium
The sum of coordination number and oxidetion number of the metal M in the complex $$\left [ M\left ( en \right )_{2}\left ( C_{2}O_{4} \right ) \right ]$$ Cl (where en is ethylenediamine) is:
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6
0%
7
0%
8
0%
9
Explanation
$$[M(en)_2(C_2O_4)]Cl$$
this can be written as $$[M(en)_2(C_2O_4)]^{+1}$$
Consider $$X$$ as oxidation number,
$$X-4-2=+1$$
$$X=+7$$
so oxidation number is $$+7$$
Which of the following are not true about hexagonal close packing?
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It has a coordination number of $$6$$
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It has $$26$$% empty space
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It is $$ABCABC......$$ type of arrangement
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It is as closely packed as body centred cubic packing
Explanation
Hexagonal closed pack has the third layer with the same arrangement of spheres as the first layer covers all the tetrahedral holes. They efficiently occupy $$74\%$$ of space while $$26\%$$ is empty. It has a co-ordination number of $$12$$ and contains $$6$$ $$atoms/unit$$ cell.
Which of the following molecules are formed by $$p-p$$ over lapping?
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$$Cl_2$$
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$$HCl$$
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$$H_2O$$
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$$NH_3$$
Explanation
Solution:- (A) $${Cl}_{2}$$
$$p-p$$ overlapping is produced by the overlapping of the $$p$$-orbitals of the atoms. In the case of chlorine molecule, it is produced by the overlapping of the $$3{p}_{z}$$ orbitals of two chlorine atoms.
In aqueous solution 1 mole of $$CoCl_3.4NH_3$$ gives 1 mole of AgCl precipitate with excess $$AgNO_3$$, then what is the secondary valency of metal?
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3
0%
4
0%
5
0%
6
Explanation
$$1$$ Mole of $$CoC{l_3}.4HN{O_3}$$ gives $$1$$ mole of $$AgCl$$
According to Alfred Werner theory of co-ordination compound the secondarty valence of metal is $$6$$
Hence, o
ption $$(D)$$ is the correct answer.
A six-coordinate complex of formula $$CrCl_3. 6H_2O $$ has a green colour. A 0.1 M solution of the complex, when treated with excess $$AgNO_3$$, gave 28.7 g of white precipitate. The formula of the complex would be:
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$$[Cr(H_2O)_6]Cl_3 $$
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$$[Cr(H_2O)_5Cl]Cl_2.H_2O$$
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Both A & B
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None of these
With a change in hybridization of the carbon bearing the charge, the stability of a carbanion decreases in the order
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$$sp < { sp }^{ 2 } < { sp }^{ 3 }$$
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$$sp < { sp }^{ 3 } < { sp }^{ 2 }$$
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$${ sp }^{ 3 } < { sp }^{ 2 } < sp$$
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$${ sp }^{ 2 } < sp < { sp }^{ 3 }$$
Explanation
Stability of carbanions increases with increase in $$s-$$ character of hybrid orbital of carbon bearing charge hence correct order is:
$$ sp^{3} < sp^{2} < sp$$
The complex ion$$\left[ Fe({ CN) }_{ 6 } \right] ^{ 4- }$$ contains :
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$${ sp }^{ 3 }{ d }^{ 2 }$$ hybrid orbitals with octahedral structure
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total 104 electrons on $${ Fe }^{ +2 }$$
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six sigma Bonds
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total of 36 electrons on $${ Fe }^{ 2+ }$$ ions.
Explanation
The hybridisation of $$[Fe(Cn)_6]_4^{3+}$$ is $$d^2sp^3$$ and the shape is octahedral due to the strong field ligand $$CN^-$$
* $$Fe^{2+}$$ has already 24 electrons in electronic configuration and $$6C\bar N$$ molecules donate 12 electrons combine gives 36 electrons on $$fe^{2+}$$ ions.
According to Werner's theory the primary valency and secondary valency in complex $$[Co(NH_3)_4Cl_2]Cl$$ is respectively:
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3,3
0%
3, 6
0%
6, 3
0%
4, 2
Explanation
The primary valency is the oxidation state of the metal ion and the coordination number of the metal is the secondary valency of the metal ion.
The oxidation number of $$Co = x + 0 -3 = 0$$
$$x = 3$$
The primary valency is 3 and the secondary valency is 6. Therefore, the correct option is B.
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