MCQ Questions for Class 12 Engineering Chemistry Coordination Compounds Quiz 8

Choose the correct statement regarding

$SeOCI_2$ molecule :

0%
It does not contain plane of symmetry
0%
$'Cl \hat{Se} Cl'$ bond angle is greater than $Cl \hat{Se} O'$ bond angle
0%
Lone pair has greater than $33.3 \%$ s-character
0%
Central atom used one d-orbital in bonding

Explanation

The structure of

$SeOCl_2$ is as follows :

Thus is does have plane of symmetry.

1798503_1816863_ans_ea2a1e4894714033b574a2c2de499b18.png

Which among the following molecules is not perfect flat?

0%
$B_3N_3H_6$
0%
$C_3N_3(NH_2)_3$
0%
$SO_3$
0%
$C_3N_3(N_3)_3$

Explanation

Option B is the answer. Its having

$N=N=N$ structure which makes it to lie in different plane. That is with respect to the base plane

$N=N=N$ lies perpendicular.

1798510_1817062_ans_c6187fc2436d41bca47ecd8c0008013e.png

Which of the following molecular species is/are having

$\pi_{2p}$ as H.O.M.O. (highest occupied molecular orbital):

0%
$N_2^-$
0%
$O_2^{2+}$
0%
$NO^+$
0%
$B^+_2$

Explanation

${O_2}^{2+}, NO^+, {B_2}^+$ have

$\pi_{2p}$ as HOMO

Least stable hybride is :

0%
stannane
0%
silane
0%
plumbane
0%
germane

Which of the following isomerism is not possible for complexes having molecular formulae?
(I)

$Pt(SCN)_{2}.3PEt_{3}$ , (II)

$CoBr . SO_{4} . 5NH_{3}$ (III)

$FeCl_{2} . 6H_{2}O$ .

0%
Optical
0%
Linkage
0%
Ionisation
0%
Hydrate

Explanation

(I)

$[Pt(SCN)(3PEt_{3})](SCN)$ : Exhibits only linkage isomerism.

as the thiocyanate group could be connected to the metal atom by either the S atom or the N atom. i.e thiocyanate,

$SCN^⁻$ and isothiocyanate,

$NCS^⁻$

(II)

$[CoBr(NH_{3})_{5}]SO_{4}$ : Exhibits only ionization isomerism.

as it can give different ion in solution i.e

$[CoBr(NH_3)_5]^{+2},SO_4^{2-} \ and\ [Co(SO_4)(NH_3)_5]^{+1},Br^-$

(III)

$[Fe(H_{2}O)_{6}]Cl_{2}$ : Exhibits only hydrate isomerism.

Hydrate isomers are the type of isomers that have. similar composition but differ in the presence of the number of water molecules. ligands.

thus, the correct answer is A.

Which of the following will have three stereoisomeric forms?
(i)

$[Cr(NO_{3})_{3}(NH_{3})_{3}]$
(ii)

$K_{3}[Co(C_{2}O_{4})_{3}]$
(iii)

$K_{3}[CoCl_{2}(C_{2}O_{4})_{2}]$
(iv)

$[CoBrCl(en)_{2}]$

0%
(iii) and (iv)
0%
(i), (iii) and (iv)
0%
(iv) only
0%
All four

Which of the following is different among structural isomers?

0%
Oxidisation state
0%
Co-ordination number
0%
IUPAC name
0%
None of these

In which of the following complex transition of electron is from one shell to other shell of central metal?

0%
$[Fe(H_{2}O)_{5}(NO)]^{2+}$
0%
$[Co(H_{2}O)_{6}]^{2+}$
0%
$[Rh(NH_{3})_{6}]^{2+}$
0%
$[Ni(CN)_{6}]^{4-}$

A metal

$M$ having electronic configuration

$(n - 1)d^{8}ns^{2}$ forms complexes with co-ordination No.

$= 4$ and

$6$ , if it forms diamagnetic complexes then permissible oxidation states of metal cation and geometry is

0%
$+2$ , octahedral
0%
$+4$ , octahedral
0%
$+2$ , square planar
0%
(b) and (c) both

Explanation

Metal, M can be Nickel.

So the permissible oxidation states of metal cation and geometry are

$+4$ , octahedral
$+2$ , square planar
$+2$ , tetrahedral

Hence, Option "D" is the correct answer.

Which of the following process is/are associated with change of hybridization of the underlined compound?

0%
$\underline {Al(OH)_3}$ ppt. dissolved in $NaOH$
0%
$\underline{B_2H_6}$ is dissolved in $THF$
0%
$\underline{SiF_4}$ vapour is passed through liq. $HF$
0%
Solidification $\underline{PCI_5}$ vapour

$[PdCl_{2}(PMe_{3})_{2}]$ is a diamagnetic complex of

$Pd(II)$ . How many total isomers are possible of analogous paramagnetic complex of

$Ni(II)$ ?

0%
Zero
0%
$1$
0%
$2$
0%
$3$

Explanation

$[NiCl_{3}(PMe_{3})_{2}]$

Hence, Option "B" is the correct answer.

Which of the following complex ion is expected to absorb light in

$4000\overset {\circ}{A}$ to

$7800\overset {\circ}{A}$ region>

0%
$[Ti(en)_{3}]^{4+}$
0%
$[Cr(H_{2}O)_{6}]^{3+}$
0%
$[Sc(NH_{3})_{4}(H_{2}O)_{2}]^{3+}$
0%
$[Zn(en)_{2}(NH_{3})_{2}]^{2+}$

Explanation

$[\overset {+IV}{Ti}(en)_{3}]^{4+}$ : No. of unpaired

$\overline {e}s = 0$

$[Sc(NH_{3})_{4}(H_{2}O)_{2}]^{3+}$ : No. of unpaired

$\overline {e}s = 0$

These complexes do not absorb light in visible range due to absence of unpaired electron and hence they are colourless.

While, in complex

$[Cr(H_{2}O)_{6}]^{3+}$ there are unpaired

$\overline {e}s$ and it is coloured due to absorption of light in visible range.

thus, the correct answer is B.

If molar conductivity of complex is almost equal to that of

$NaCl$ and it does not exhibits stereoisomerism then the complex will be

0%
$[Co(CO_{3})(en)_{2}]Br$
0%
$[Co(CO_{3})(H_{2}O)_{2}(NH_{3})_{2}]Br$
0%
$[Co(CN)(NH_{3})_{5}]Br_{2}$
0%
$[Co(CO_{3})(NH_{3})_{4}]Br$

Explanation

The complex

$[Co(CO_{3})(NH_{3})_{4}]Br$ has a molar conductivity similar to as that of NaCl and it also does not show stereo Isomerism.

Hence, Option "D" is the correct answer.

Which of the following orbitals can not undergo hybridisation amongst themselves.
(I) 3d, 4s (II) 3d, 4d
(III) 3d,4s and 4p (IV) 3s, 3p and 4s

0%
only II
0%
II and III
0%
I, II and IV
0%
II and IV

Consider the following isomers
(i)

$[Pt(NH_{3})_{4}Cl_{2}]Br$
(ii)

$[Pt(NH_{3})_{4}Br_{2}]Cl_{2}$
(iii)

$[Co(NH_{3})_{4}Cl_{2}]NO_{2}$

Which of the following observation is correct?

0%
(i) will give a pale yellow and (ii) will give a white precipitate with $AgNO_{3}$ solution.
0%
(iii) will give a white precipitate with $AgNO_{3}$ solution
0%
(i), (ii) and (iii) will give white precipitate with $AgNO_{3}$ solution.
0%
None of the above isomers will give white precipitate with $AgNO_{3}$ solution.

Explanation

$\bullet$

$\mathrm{[Pt(NH_{3})_{4}Cl_{2}]Br_{2}\rightarrow [Pt(NH_{3})_{4}Cl_{2}]^{2+}+2Br^{-}}$

$\mathrm{Br^{-}+AgNO_{3}\rightarrow \underset{pale\ yellow}{AgBr}+NO_{3}^{-}}$

$\bullet$

$\mathrm{[Pt(NH_{3})_{4}Br_{2}]Cl_{2}\rightarrow [Pt(NH_{3})_{4}Br_{2}]^{2+}+2Cl^{-}}$

$\mathrm{Cl^{-}+AgNO_{3}\rightarrow \underset{white}{AgCl}+NO_{3}^{-}}$

$\bullet$

$\mathrm{[Co(NH_3)_4Cl_2]NO_2 \rightarrow [Co(NH_3)_4Cl_2]^+ + NO_2}^-$

$\mathrm{NO_2^- + AgNO_3 \rightarrow \underset{Water \ Soluble}{AgNO_2}}$

Hence, Option "A" is the correct answer.

Types of isomerism exhibited by

$[CrCl_{2}(NO_{2})_{2}(NH_{3})_{2}]^{-}$ complex ion are

0%
Ionisation, optical
0%
Hydrate, optical
0%
Geometrical, optical
0%
Co-ordinate, geometrical

Which of the following will not show geometrical isomerism?

0%
$[Cr(NH_{3})_{4}Cl_{2}]Cl$
0%
$[Co(en)_{2}Cl_{2}]Cl$
0%
$[Co(NH_{3})_{5}NO_{2}]Cl_{2}$
0%
$[Pt(NH_{3})_{2}Cl_{2}]$

Explanation

Octahedral complex of type

$\mathrm{[MA_{5}B] \ \ \& \ \ [MA_6]}$ cannot show geometrical isomerism.

Here,

$\mathrm{[Co(NH_3)_5NO_2]Cl}$ is of

$\mathrm{[MA_5B]}$ type. So, it does not show geometrical isomerism.

Hence, Option "C" is the correct answer.

Both geometrical and optical isomerism are shown by

0%
$[Co(en)_{2}Cl_{2}]^{+}$
0%
$[Co(NH_{3})_{5}Cl]^{2+}$
0%
$[Co(NH_{3})_{4}Cl_{2}]^{+}$
0%
$[Co(ox)_{3}]^{3-}$

Explanation

Octahedral complexes of type

$\mathrm{[M(\widehat{aa}) B_2]}$ , show both geometrical as well as optical isomerism.

Of all given options,

$\mathrm{[Co(en)Cl_2]^+}$ is of form

$\mathrm{[M(\widehat{aa}) B_2]}$ .

Hence, Option "A" is the correct answer.

One mole of a complex compound

$Co(NH_{3})_{5}Cl_{3}$ gives three moles of ions on dissolution in water. One mole of the same complex reacts with two moles of

$AgNO_{3}$ and yields two moles of

$AgCl{(s)}$ . The complex is:

0%
$[Co(NH_{3})_{4}Cl_{2}]$ $Cl.NH_{3}$
0%
$[Co(NH_{3})_{4}Cl]$ $Cl_{2}.NH_{3}$
0%
$[Co(NH_{3})_{5}Cl]$ $Cl_{2}$
0%
$[Co(NH_{3})_{3}Cl]$ $Cl_{3}.2NH_{3}$

Explanation

It is given that one mole of a complex compound

$Co(NH_{3})_{5}Cl_{3}$ gives three moles of ions on dissolution in water. One mole of the same complex reacts with two moles of

$AgNO_{3}$ and yields two moles of

$AgCl{(s)}$ .

The complex ion is

$[Co(NH_{3})_{5}Cl]$

$Cl_{2}$ as two

$Cl$ ions are out of the coordination complex.

Therefore, on dissolution in water, three ions are produced and they are

$2Cl^-$ and

$[Co(NH_3)_5Cl]^{2+}$ ions.

The two

$Cl^-$ ions which are free react with

$2$ moles of

$AgNO_3$ to give

$2$ moles of

$AgCl$ .

Mixture

$[X] = 0.02$ mole of

$[Co(NH_3)_5SO_4]Br$ and 0.02 mol of

$[Co(NH_3)_5Br]SO_4$ was prepared in 2 litre of solution.
1 litre of mixture [X] + excess

$AgNO_3 \longrightarrow$ [Y]
1 litre of mixture [X] + excess

$BaCl_2\longrightarrow$ [Z]
No. of moles of [Y] and [Z] are :

0%
$0.01, 0.01$
0%
$0.02, 0.01$
0%
$0.01, 0.02$
0%
$0.02,0.02$

Explanation

Total volume of the solution is 2 L,

Mixture X contains 0.02 moles of

$[Co(NH_3)_5 SO_4] Br$

Therefore Molarity of

$[Co(NH_3)_5SO_4 ]Br,$

$\cfrac{\text {No. of moles of solute}} {\text {Volume of solution in Litres}} = \cfrac{0.02} {2} = 0.01 M$

Mixture of X also contains 0.02 moles of

$[Co(NH_3)_5 Br] SO_4,$

Therefore, Molarity of

$[Co(NH_3)_5 Br] SO_4=$

$\cfrac{\text {No. of moles of solute}} {\text {Volume of solution in Litres}} = \cfrac{0.02} {2} = 0.01 M$

Now, according to the reaction,

$\underset{(0.01M)}{[Co(NH_3)_5 SO_4] Br} + AgNO_3 \rightarrow [Co(NH_3)_5 SO_4] NO_3 (Soluble) +\underset{[Y]ppt}{ AgBr} \downarrow$

According to Avogadro law, the molarity of

$AgBr$ will also be 0.01, since molarity is moles per litre. Therefore, the number of moles of

$AgBr$ will also be 0.01.

Similarly in the second reaction,

$BaSO_4$ (Z) will be precipitated and

$[Co(NH_3)_5 Br] Cl_2$ will be formed which is soluble. Therefore, no. of moles of

$BaSO_4$ = 0.01

The solubility of silver bromide in hypo solution is due to the formation of:

0%
$Ag_{2}SO_{3}$
0%
$Ag_{2}S_{2}O_{3}$
0%
$[Ag(S_{2}O_{3})]^{-}$
0%
$[Ag(S_{2}O_{3})_{2}]^{3-}$

Explanation

Unreacted silver bromide

$(AgBr)$ can be dissolved from photographic emulsions using 'hypo' solution

$(Na_2S_2O_3)$ (aq.). The reaction can be represented as:

$AgBr(s)+3Na_2S_2O_3(aq)\rightarrow Na_5Ag(S_2O_3)_3(aq)+NaBr(aq)$

Hence, option

$D$ is correct.

Which of the following will give maximum number of isomers?

0%
$\left [ Co\left ( NH_{3} \right )_{4} Cl_{2}\right ]$
0%
$\left [ Ni\left ( en \right )\left ( NH_{3} \right )_{4} \right ]^{2+}$
0%
$\left [ Ni\left ( C_{2}O_{4} \right )\left ( en \right )_{2} \right ]$
0%
$\left [ Cr\left ( SCN \right )_{2}\left ( NH_{3} \right )_{4} \right ]^{2+}$

Explanation

$[Cr(SCN)_2(NH_3)_4]^{2+}$ has the maximum number of isomers (geometrical + linkage).

$\left [ Ni\left ( en \right )\left ( NH_{3} \right )_{4} \right ]^{2+}$ ,

$\left [ Ni\left ( C_{2}O_{4} \right )\left ( en \right )_{2} \right ]$ and

$\left [ Co\left ( NH_{3} \right )_{4} Cl_{2}\right ]$ have 2 geometrical isomers each.

Which of the following ions is expected to be coloured?

0%
$\left [ Ti\left ( H_{2}O \right )_{6} \right ]^{3+}$
0%
$\left [ Sc\left ( H_{2}O \right )_{6} \right ]^{3+}$
0%
$\left [ Zn\left ( H_{2}O \right )_{6} \right ]^{2+}$
0%
$\left [ Ti\left ( H_{2}O \right )_{6} \right ]^{4+}$

Explanation

A) Ti has the electronic configuration

$[Ar] 4s^23d^2$ . In this complex, it is present as

$Ti^{3+}$ which has one unpaired electron in 3d orbital. It undergoes

$d^2sp^3$ (octahedral) hybridization to accommodate 6 pairs of water.
So, the complex is octahedral in shape, will be coloured and paramagnetic due to the presence of one unpaired electron.

B) Sc has the electronic configuration

$[Ar] 4s^23d^1$ . Since, it is present in the complex as

$Sc^{3+}$ , it does not have any unpaired electron. It undergoes

$d^2sp^3$ hybridization to accommodate 6 pairs of electrons from water. So, it is octahedral in shape, colourless and diamagnetic due to the absence of unpaired electrons.

C) In this complex,

$Zn$ is present as

$Zn^{2+}$ and there is no unpaired electron in the complex. So, it is colourless.

D) In this complex,

$Ti$ is present as

$Ti^{4+}$ and there is no unpaired electron in the complex. So, it is colourless.

The observed magnetic moment value $(\mu _{abs})$ is higher than calculated magnetic moment value $(\mu _{cal})$ for which of the following species?

0%
Ti $^{+3}$
0%
V $^{+2}$
0%
Co $^{+2}$
0%
Cr $^{+2}$

Explanation

The observed magnetic moment value

$(\mu _{abs})$ of

$Co^{2+}$ is in the range of

$4.3-5.2$ which is higher than calculated magnetic moment value

$(\mu _{cal})$ which is

$3.87$ . Values are different enough that they usually confirm the number of unpaired spins (S) and indicate whether the complex is high spin or low spin.

Match list-I (Cooordination Compounds) with list-II (Type of Isomerism) and select the correct answer using the codes given below :

	List-I		List-II
A.	$Co(NH_3)_4Cl_2$	1.	Optical isomerism
B.	$Cis-Co(en)_3Cl_3$	2.	Ionization Isomerism
C.	$Co(en)_2(NH_3)Cl_2SCN$	3.	Co-ordination Isomerism
D.	$Co(NH_3)_6Cr(CN)_6$	4.	Geometrical Isomerism

0%
A-4, B-3, C-2, D-1
0%
A-1, B-4, C-3, D-2
0%
A-4, B-1, C-2, D-3
0%
A-4, B-2, C-3, D-1

Explanation

$[Co(NH_3)_4Cl_2]$ shows geometrical isomerism. It exists in cis and trans form.

$Cis- [Co(en)_3]Cl_3$ shows optical isomerism. It exists in

$d-$ and

$l-$ form.

$[Co(en)_2(NH_3)Cl_2SCN$ exhibits ionization isomerism.

$[Co(NH_3)_6][Cr(CN)_6]$ exhibits coordination isomerism.

The slight irregularities in size have been ascribed to:

0%
crystal field effects
0%
shapes of d orbitals
0%
charge transfer
0%
splitting of d orbitals

The total number of possible coordination isomers of the complex compound

$\left [ Cu\left ( NH_{3} \right )_{4} \right ]\left [ PtCl_{4} \right ]$ are:

0%
3
0%
6
0%
2
0%
4

Explanation

The total number of possible isomers of the complex compound

$\left [ Cu\left ( NH_{3} \right )_{4} \right ]\left [ PtCl_{4} \right ]$ are 4.

$\left [ Cu\left ( NH_{3} \right )_{4} \right ]\left [ PtCl_{4} \right ]$

$\left [ Cu\left ( NH_{3} \right )_{3}Cl \right ]\left [ Pt\left ( NH_{3} \right )Cl_{3} \right ]$

$\left [ Pt\left ( NH_{3} \right )_{3}Cl \right ]\left [ Cu\left ( NH_{3} \right ) Cl_{3} \right ]$

$\left [ Pt(NH_3)_{4} \right ]\left [ Cu\left (Cl \right )_{4} \right ]$

Option D is correct.

A complex compound of cobalt has molecular formula containing five

$NH_{3}$ molecules, one

$NO_{2}$ group and two

$Cl$ atoms for one

$Co$ atom. One mole of this compound produces three moles of ions in aqueous solution. On reacting with an excess of

$AgNO_{3}$ solution, two moles of

$AgCl$ get precipitated. The ionic formula of the compound is:

0%
$[Co(NH_{3})_{4} NO_{2}Cl]$ $[(NH_{3})Cl]$
0%
$[Co(NH_{3})_{5} Cl]$ $[Cl (NO_{2})]$
0%
$[Co (NH_{3})_{5} (NO_{2})] Cl_{2}$
0%
$[Co(NH_{3})_{5}] [(NO_{2}) Cl_{3}]$

Explanation

The ionic formula of the compound is

$[Co (NH_{3})_{5} (NO_{2})] Cl_{2}$ .

$[Co (NH_{3})_{5} (NO_{2})] Cl_{2} \rightarrow [Co (NH_{3})_{5} (NO_{2})]^+ + 2Cl^{-}$

Total 3 ions are produced in aqueous solution.

Hence, the correct option is

$\text{C}$

Which pair is correctly matched?

0%
$[Fe(OH)_3] Fe^{3+} : Cl^{-}$
0%
$[As_{2}S_{3} ]: As^{3+}$
0%
$[\mathrm{S}\mathrm{n}\mathrm{O}_{2}]:\mathrm{S}\mathrm{n}\mathrm{O}_{3}^{2-}$ in acidic medium
0%
$[\mathrm{A}\mathrm{g}\mathrm{I}]:\mathrm{I}^{-}$ in excess of $\mathrm{A}\mathrm{g}\mathrm{N}\mathrm{O}_3$

Which one of the following has largest number of isomers? (

$\mathrm{R}=$ alkyl group, en

$=$ ethylenediamine)

0%
$[\mathrm{R}\mathrm{u}(\mathrm{N}\mathrm{H}_{3})_{4}\mathrm{C}1_{2}]^{+}$
0%
$[\mathrm{C}\mathrm{o}(\mathrm{N}\mathrm{H}_{3})_{5}\mathrm{C}1]^{2+}$
0%
$[\mathrm{I}\mathrm{r}(\mathrm{P}\mathrm{R}_{3})_{2}\mathrm{H}(\mathrm{C}\mathrm{O})]^{2+}$
0%
$[\mathrm{C}\mathrm{o}(\mathrm{e}\mathrm{n})_{2}\mathrm{C}1_{2}]^{+}$

Explanation

$[\mathrm{C}\mathrm{o}(\mathrm{e}\mathrm{n})_{2}\mathrm{C}1_{2}]^{+}$ has largest number of isomers which includes geometrical isomers (cis and trans isomers) as well as stereoisomers (d form, l form and optically inactive form).

$en$ is a bidentate ligand.

The correct statement in respect of protein haemoglobin is that it :-

0%
Maintains blood sugar level
0%
Acts as an oxygen carrier in the blood
0%
Forms antibodies and offers resistance to diseases
0%
Functions as a catalyst for biological reactions

In which of the following coordination entities the magnitude of

$\Delta _{0}$ (CFSE in the octahedral field) will be maximum? (Atomic number of

$Co=27$ )

0%
$[Co(CN)_{6}]^{3-}$
0%
$[Co(C_{2}O_{4})_{3}]^{3-}$
0%
$[Co(H_{2}O)_{6}]^{3+}$
0%
$[Co(NH_{3})_{6}]^{3+}$

Which one of the following has largest number of stereoisomers?

0%
$[Ru(NH_{3})_{4}Cl_{2}]^{+}$
0%
$[Co(NH_{3})_{5}Cl]^{2+}$
0%
$[Ir(PR_{3})_{2}H(CO)]^{2+}$
0%
$[Co(en)_{2}Cl_{2}]^{+}$

The pair(s) of coordination complexes/ions exhibiting the same kind of isomerism is(are):

0%
$[Cr(NH_{3})_{5}Cl]Cl_{2}$ and $[Cr(NH_{3})_{4}Cl_{2}]Cl$
0%
$[Co(NH_{3})_{4}Cl_{2}]^{+}$ and $[Pt(NH_{3})_{2}(H_{2}O)Cl]^{+}$
0%
$[CoBr_{2}Cl_{2}]^{2-}$ and $[PtBr_{2}Cl_{2}]^{2-}$
0%
$[Pt(NH_{3})_{3}(NO_{3})] Cl$ and $[Pt(NH_{3})_{3}Cl]Br$

Explanation

The pairs of coordination complexes/ions exhibiting the same kind of isomerism are: Octahedral

$[Co(NH_{3})_{4}Cl_{2}]^{+}$ and square planar

$[Pt(NH_{3})_{2}(H_{2}O)Cl]^{+}$ which shows geometrical (cis-trans) isomerism.

Square planar

$[Pt(NH_{3})_{3}(NO_{3})] Cl$ and square planar

$[Pt(NH_{3})_{3}Cl]Br$ shows ionization isomerism.

The pairs of coordination complexes/ions NOT exhibiting the same kind of isomerism are:

$\displaystyle [Cr(NH_{3})_{5}Cl]Cl_{2}$ neither shows structural nor stereoisomerism.

$\displaystyle [Cr(NH_{3})_{4}Cl_{2}]Cl$ shows geometrical (cis trans) isomerism.

Tetrahedral

$\displaystyle [CoBr_{2}Cl_{2}]^{2-}$ and square planar

$\displaystyle [PtBr_{2}Cl_{2}]^{2-}$ do not show same type of isomerism.

Hence, options

$B$ and

$D$ are correct.

The ratio of pure and hybrid orbitals of valence shell in

$H_{2}C=CH-CH=CH_{2}$ is:

0%
$7 : 12$
0%
$14 : 13$
0%
$6 : 5$
0%
$5 : 6$

Explanation

Every double bonded carbon has

$3$

$sp^2$ orbitals and one pure p orbital. And every hydrogen has one pure s orbital, so the number of hybrid orbitals are

$4 \times 3=12$ , and pure orbitals are

$1 \times 4+6 \times 1=10$ .
So, their ratio is

$10:12$ or

$5:6$ .

In the analysis of an organic compound, it was found that it contains 7 carbon atoms, there are two C = C bonds and one

$C\equiv C$ bond. This compound is a hydrocarbon. Hydrocarbons have only carbon and hydrogen elements. On structural analysis it was found that it is covalent in nature and expected structure is given above. The ratio between the pure and hybrid orbitals is?

0%
$7:10$
0%
$8:7$
0%
$5:9$
0%
$1:2$

Explanation

$\displaystyle sp, sp^2$ and

$\displaystyle sp^3$ hybridization of

$C$ , the number of hybrid orbitals formed are

$2,3$ and

$4$ respectively and the number of pure orbitals are

$2,1$ and

$0$ respectively.
In this compound,

$2$

$\displaystyle sp$ hybridized,

$4$

$\displaystyle sp^2$ hybridized and

$1$

$\displaystyle sp^3$ hybridized

$C$ atom is present. The number of pure orbitals in these atoms are

$\displaystyle 4+4+0=8$
Also there are

$6\: H$ atoms which contribute

$6$ pure orbitals. Hence,the total number of pure orbitals are

$\displaystyle 8+6=14$
The number of hybrid orbitals are

$\displaystyle 4+12+4+0=20$
Hence, the ratio of pure orbitals to hybrid orbitals is

$\displaystyle 14: 20$ or

$\displaystyle 7:10$

Match the correct pairs of isomers with their examples.

A)Ionisation (a) $Pt(NH_{3})_{2}Cl_{2}][Pt(NH_{3})_{4}]$

$[PtCl_{4}]$

B) Linkage (b) $[Cr(NH_{3})_{6}][Co(CN)_{6}]$ & $[Co(NH_{3})_{6}[Cr(CN)_{6}]$

C)Coordination (c) $[Co(NH_{3})_{5}(NO_{2})]Cl_{2}$ & $[Co(NH_{3})_{5}(ONO)]Cl_{2}$

D)Polymerisation (d) $[Co(SO_{4})(NH_{3})_{5}]Br$ & $[Co(Br (NH_{3})_{5}]SO_{4}$

0%
A -> a, B -> b, C -> c, D -> d
0%
A -> b, B -> a, C -> d, D -> c
0%
A -> d, B -> c, C -> b, D -> a
0%
A -> d, B -> b, C -> c, D -> a

Explanation

$Polymerization\, Isomerism$ : This is not true isomerism because it occurs between compounds having the same empirical formula, but different molecular weights.

For example,

$[Pt(NH_{3})_{2}CI_{2}],[Pt(NH_{3})_{4}][PtCI_{4}],[Pt(NH_{3})_{4}][Pt(NH_{3})CI_{3}]_{2}$ .

$Inonization\, Isomerism$ : This type of isomerism is due to the exchange of groups between the complex ion and the ions outside it.

$[Co(NH_{3})_{5}Br]SO_{4}$ is red-violet. An aqueous solution gives a white precipitate of

$BaSO_{4}$ with

$BaCI_{2}$ solution, thus confirming the presence of free

$SO_{4}^{2-}$ ions. In contrast

$[Co(NH_{3})_{5}SO_{4}]Br$ is red. A solution of this complex does not give a positive sulphate test with

$BaCI_{2}$ .It does give a cream-coloured precipitate of

$AgBr$ with

$AgNO_{3}$ , thus confirming the presence of free

$Br^{-}$ ions.

$Linkage\, Isomerism$ : Certain ligands contain more than one atom which could donate an electron pair. In the

$NO_{2}^{-}$ group in the complex ion. In the

$NO^-_2$ ion, either

$N$ or

$O$ atoms could act as the electrons pair donor. Thus there is the possibility of isomerism. Two different complexes

$[Co(NH_3)_5NO)_2]Cl_2$ and

$[Co(NH_3)_5(ONO)]Cl_2$ have been prepared, each containing the

$NO^-_2$ group in the complex ion.

$Coordination\,Isomerism$ : If the complex is a salt having both cation and anion as complex ions then the ligands can exchange position between the cation and the anion. This will result in the formation of coordinates isomers. For example

$[Co(en)_{3}][Cr(C_{2}O_{4})_{3}]$ and

$[Co(en)_{2}(C_{2}O_{4})][Cr(en)(C_{2}O_{4})_{2}]$

$[Cr(en)_{2}(C_{2}O_{4})][Co(en)(C_{2}O_{4})_{2}]$ and

$[Cr(en)_{3}][Co(C_{2}O_{4})_{3}]$

Which of the following statements is incorrect?

0%
Complex $[Co(NH_3)_4(H_2O)Cl]Br_2$ can show both hydrate as well as ionization isomerism
0%
Complex $[Co(NH_3)_5(H_2O)](NO_3)_3$ can show hydrate isomerism
0%
Complex $[Pt(NH_3)_4][PtCl_6]$ cannot show coordination isomerism
0%
$[Co(NH_3)_4(NO_2)Cl]Cl$ can show both ionization as well as linkage isomerism

Explanation

Option (C) is correct.

(A) :

$[Co(NH_3)_4(H_2O)Cl]Br_2$ and

$[Co(NH_3)_4Br_2]Cl.H_2O$ ; ionization as well as hydrate isomers.
(B) :

$[Co(NH_3)_5(H_2O)](NO_3)_3$ and

$[Co(NH_3)_5(NO_3)](NO_3)_2H_2O$ ; hydrate isomers.
(C) :

$[Pt(NH_3)_4]^{2+} [PtCl_6]^{2-}$ and

$[Pt(NH_3)_4Cl_2]^{2+} [PtCl_4]^{2-}$ ; coordination isomers.
(D) :

$[Co(NH_3)_4(NO_2)Cl]Cl$ ;

$[Co(NH_3)_4(ONO)Cl]Cl\!^-$ ; linkage isomer and

$[Co(NH_3)_4Cl_2]{NO_{2}}^{-}$ ; ionization isomer.

In which of the following complexes more than one type of structural isomerism is possible?

0%
$[Co(NH_{3})_{5} (NO_{2})]Cl$
0%
$[Co(NH_{3})_{5}(H_{2}O)](NO_{2})_{3}$
0%
$[Pt(NH_{3})_{4}][Pt(SCN)_{4}]$
0%
$[Cr(DMG)_{3}]$

Explanation

Structural isomerism in coordination compounds can be of following types-
Ionisation isomerism

This is due to difference in ionisable group,
Eg:

$\left[ CO{ \left( N{ H }_{ 3 } \right) }_{ 5 }Cl \right] S{ O }_{ 4 }$ &

$\left[ CO{ \left( N{ H }_{ 3 } \right) }_{ 5 }S{ O }_{ 4 } \right] Cl$
Polymerisation isomerism

Compounds having same emperical formula, but different molecular weights.
Eg-

$\left[ Pt{ \left( N{ H }_{ 3 } \right) }_{ 2 }Cl \right]$ &

$\left[ Pt{ \left( N{ H }_{ 3 } \right) }_{ 4 }Cl \right] \left[ Pt{ Cl }_{ 4 } \right]$
Hydrate isomerism

Isomerism due to different number of water molecules attached to ligands.
Eg:

$\left[ Cr{ \left( { H }_{ 2 }O \right) }_{ 6 } \right] { Cl }_{ 3 }$ &

${ \left[ Cr{ \left( { H }_{ 2 }O \right) }_{ 5 }Cl \right] }{ Cl }_{ 2 }({ H }_{ 2 }O)$
Linkage Isomerim

Having different kind of linkage to the central metal ion.
Coordination Isomerism

When ligand itself exhibits different isomeric forms.
Eg:

$\left[ CO{ \left( N{ H }_{ 3 } \right) }_{ 6 } \right] \left[ Cr{ \left( CN \right) }_{ 6 } \right]$ &

$\left[ Cr{ \left( N{ H }_{ 3 } \right) }_{ 6 } \right] \left[ CO{ \left( CN \right) }_{ 6 } \right]$

Ligand Isomerism

When ligand itself exhibits different isomeric forms
Eg:

$O-toluidine$ &

$m-toluidine$ .
Only option (

$D$ ) couldn't exhibit more than one type of isomerism, because of the same three ligands (

$DMG$ ).

The total number of isomers shown by

$[Co(NH_3)_4(NO_2)_2](NO_3)$ complex is:

0%
10
0%
6
0%
4
0%
12

Explanation

The complex

$[Co(NH_3)_4(NO_2)_2](NO_3)]$ has ten isomers. They include geometrical, ionization and linkage isomers.

$NO_3$ and

$NO_2$ groups can be either inside or outside the coordination sphere. This gives ionization isomers.

$NO_2$ group can coordinate through N or O atom. This gives linkage isomerism. cis and trans isomers give geometrical isomers.

$[Fe(H_2O)_4(CN)_2]$ is the empirical formula of a compound which has a magnetic moment corresponding to

$2\frac{2}{3}$ unpaired electrons per iron. The best possible formula of the compound is

0%
$[Fe(H_2O)_6]2[Fe(CN)_6]$
0%
$[Fe(H_2O_6)][Fe(H_2O)_2(CN)_4]$
0%
$[Fe(H_2O)_4(CN)_2]_2[Fe(H_2O)_4(CN)_2]$
0%
None of these

Explanation

Magnetic moment

$M=\sqrt {n(n+2)}$ magneton And here

$M=2\frac {2}{3}=\frac {8}{3}$ unpaired

$e^-$ per iron.
best formula of compound is

$[Fe(H_2O)_6]_2[Fe(CN)_6]$

Match column-I (chemical composition of precipitates) with column-II (colour of the precipitate) and select the correct answer using the codes given below in the columns :
(a)

$Co [Hg (SCN)_4]$ (p) black
(b)

$Fe_3[Fe(CN)_6]_2$ (q) rosy red
(c)

$Ni(dmg)_2$ (r) prussian blue

(d) PbS (s) deep blue

0%
a-s , b-p , c-q , d-r
0%
a-s , b-r , c-q , d-p
0%
a-s , b-p , c-r , d-q
0%
a-p , b-r , c-q , d-s
0%
a-p , b-q , c-r , d-s

Explanation

$Co[Hg(SCN)_4]$ - Deep Blue

$Fe_3[Fe(CN_)6]_2$ - Prussian Blue

$Ni(dmg)_2$ - Rosy red

$PbS$ - Black

$K: K_4[Mn(CN)_6], \\L: K_3[Co(NO_2)_6], \\ M: [Co(H_2O)_6]Cl_3, \\ N:[Cr(CO)_6], \\ O:[Ni(en)_3](NO_3)_3$

$P:[Pt(NH_3)_2Cl_2]$

The diamagnetic complexes are :

0%
L, M, N, P
0%
K, L, N, O
0%
L, M, O, P
0%
K, N, O, P

Explanation

The complex K having structure

$K_4[Mn(CN)_6]$ has

$3d^5$ electron configuration. It undergoes

$d^2sp^3$ hybridization. It contains one unpaired electron and is paramagnetic.
The complex L having structure

$K_3[Co(NO_2)_6]$ has

$3d^6$ electron configuration. It undergoes

$d^2sp^3$ hybridization. It contains no unpaired electron and is diamagnetic.
The complex M having structure

$[Co(H_2O)_6]Cl_3$ has

$3d^6$ electron configuration. It undergoes

$d^2sp^3$ hybridisation. It contains no unpaired electron and is diamagnetic.
The complex N having structure

$[Cr(CO)6]$ has

$3d^6$ electron configuration. It undergoes

$d^2sp^3$ hybridization. It contains no unpaired electron and is diamagnetic.
The complex O having structure

$[Ni(en)_3](NO_3)_3$ has

$3d^8$ electron configuration. It undergoes

$sp^3d^2$ hybridization. It contains two unpaired electron and is paramagnetic.
The complex P having structure

$[Pt(NH_3)_2Cl_2]$ has

$5d^8$ electron configuration. It undergoes

$dsp^3$ hybridization. It contains no unpaired electron and is diamagnetic.

Which one of the following complexes is not expected to exhibit isomerism?

0%
$[Pt(NH_3)_2Cl_2]$
0%
$[Ni(NH_3)_2Cl_2]$
0%
$[Ni(en)_3]^{2+}$
0%
$[Ni(NH_3)_4(H_2O)_2]^{2+}$

Which of the following orbitals can not undergo hybridisation amongst themselves?

$(I)\ 3d, 4s$

$(II)\ 3d, 4d$

$(III)\ 3d, 4s$ &

$4p$

$(IV) \ 3s, 3p$ &

$4s$

0%
$II$
0%
$II$ and $III$
0%
$I, II$ and $IV$
0%
$II$ and $IV$

Explanation

$(II) \ 3d$ and

$4d$ can not undergo hybridization without involving

$4s$ .

$(IV) \ 3s, 3p$ and

$4s$ can not undergo hybridization without involving

$3d$ .

Coordination number of Ni in

$[Ni(C_2O_4)_3]^{4-}$ is:

0%
3
0%
6
0%
4
0%
2

Explanation

Co-ordination number of

$Ni$ in

$[Ni(C_2O_4)]^{-4}$ is

$6.$ . Because

$C_2O_4^{-3}$ is a bidentate ligand and these ligands are

$3$ in number. So

$2 \times 3=6$ be the co-ordination number.

Which of the following are paramagnetic?

0%
$[Ni(CN)_4]^{2-}$
0%
$[NiCl_4]^{2-}$
0%
$[CoF_6]^{3-}$
0%
$[Co(NH_3)_6]^{3+}$

Explanation

The approach to find whether a coordination complex is paramagnetic or not goes through determining the unpaired electron in the orbital.

We know very well that if the ligand in the complex is strong then it can pair up the electrons in the low energy orbital first and then the electrons are filled up in the higher energy orbital.

Let's check the given options one by one-

The electronic configuration of Ni is-

Ni(28) -

$[Ar] 4s^2 3d^8$

$Ni^{2+}(26)$ -

$[Ar] 4s^0 3d^8$

(A)

$[Ni(CN)_4]^{2-} \Rightarrow CN^-$ is a strong field ligand it will pair up the electrons in the

$e_g \text {(lower energy orbital in tetrahedral complexes as shown in the given diagram)}$ orbital first and then go to the higher energy orbital

$t_{2g}$ . Since Ni has even no. of electrons in the d-orbital after pairing it will show

$\text {diamagnetism}$ means no unpaired electron. In the given diagram for tetrahedral complexes (

$T_d$ ) only

$d_{XZ}$ will remain empty rest orbitals will be fully filled.

(B)

$Ni(Cl)_4]^{2-} \Rightarrow \text {tetrahedral complex, weak field ligand}$ No pairing takes place hence all the

$t_{2g}$ and

$e_g$ orbitals will be filled with single electron first and then pairing takes place.

$d_{YZ}$ and

$d_{XZ}$ will be half filled. Hence

$\text {paramagnetic}$

Co (27) -

$[Ar] 4s^2 3d^7$

$Co^{3+} (24)$ -

$[Ar] 4s^0 3d^6$

(C)

$[CoF_6]^{3−}\Rightarrow \text {Octahedral complex, weak field ligand}$ , No pairing takes place

$d_{X^2-Y^2}$ ,

$d_{Z^2}$ ,

$d_{YZ}$ and

$d_{XZ}$ will be half filled. Hence

$\text {Paramagnetic}$ .

(D)

$[Co(NH_3)_6]^{3+}\Rightarrow \text {Octahedral complex, strong field ligand}$ , Pairing takes place.

$t_{2g}$ orbitals will be fully filled but

$e_g$ orbitals will be empty. Hence

$\text {Diamagnetic}$

Option B and C are correct.

1957885_116404_ans_6b217f79b21a4df585e7a83cc1c5e362.png

For the dark green compound,

$CrCl_6H_2O$ , which of the following is correct?

0%
all the $Cl$ shows primary valency (PV).
0%
two $Cl$ shows primary valency (PV) and one $Cl$ secondary valency (SV).
0%
two $Cl$ shows primary valency (PV) and one $Cl$ (PV) as well as secondary valency (SV).
0%
all the Cl show secondary valency (SV).

Explanation

All the Cl shows primary valency (PV). All

$\displaystyle H_2O$ show secondary valency (SV). The complex can be written as

$\displaystyle [Cr(H_2O)_6]Cl_3$ .

$\bf{Note:}$ Primary valency is the number of negative ions which are equivalent to the charge on the metal ion. Secondary valency is the number of ligands that are attached or coordinated to the metal ion.

Which of the following statement is true for the complex,

$[Co(NH_3)_4Br_2]NO_2?$

0%
It shows ionisation, linkages and geometrical isomerism.
0%
It does not show optical isomerism because its cis and trans forms each have at least one plane of symmetry.
0%
Its ionisation isomers cannot be differentiated by silver nitrate solution.
0%
(A) and (B) both.

Explanation

The complex

$[Co(NH_3)_4Br_2]NO_2$ shows geometrical isomerism but not optical isomerism as it contains at least one plane of symmetry.
The linkage isomers are

$[Co(NH_3)_4BrNO_2]Br and [Co(NH_3)_4BrONO]Br$ The ionization isomers are

$[Co(NH_3)_4Br_2]NO2$ and

$[Co(NH_3)_4BrNO_2]Br$ . The action of

$AgNO_3$ solution on

$[Co(NH_3)_4BrNO_2]Br$ gives pale yellow precipitate.

Primary and secondary valency of

$Pt$ in

$[Pt(en)_2Cl_2]Cl_2$ are:

0%
2, 4
0%
2, 6
0%
6, 6
0%
4, 2

Explanation

Primary valency is the number of negative ions attached to the central metal atom.

Secondary valency is the number of ions or molecules that are coordinated to the central metal ion; its co-ordination number

Platinum is bonded with two types of ligands,

Ethylenediamine which is a bidentate ligand is two in number hence $2\times2$ =4

Chloride which is an unindented ligand is two in number hence $1\times2$ = 2

Co-ordination no.= 6

Oxidation state of central metal atom : $x+ 0-2$ = $x$ = 2

CBSE Questions for Class 12 Engineering Chemistry Coordination Compounds Quiz 8 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 12 Engineering Chemistry Coordination Compounds Quiz 8 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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