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CHXII09:COORDINATION COMPOUNDS

322285 Which one of the following is an inner orbital complex as well as diamagnetic in behaviour? (Atomic number: $Zn = 30, Cr = 24, Co = 27, Ni$ $= 28$ )

1

{[Zn {({NH}_{3})}_{6}]}^{2 +}

2

{[Cr {({NH}_{3})}_{6}]}^{3 +}

3

{[Co {({NH}_{3})}_{6}]}^{3 +}

4

{[Ni {({NH}_{3})}_{6}]}^{2 +}

Explanation:

${[Co {({NH}_{3})}_{6}]}^{3 +}, {Co}^{3 +} - [Ar] 3 d^{6}$
supporting img
${[Cr {({NH}_{3})}_{6}]}^{3 +}, {Cr}^{3 +} - [Ar] 3 d^{3}$

${[Ni {({NH}_{3})}_{6}]}^{2 +}, {Ni}^{2 +} - [Ar] 3 d^{8}$

${Zn {({NH}_{3})}_{6}]}^{2 +}, {Zn}^{2 +} - [Ar] 3 d^{10}$

CHXII09:COORDINATION COMPOUNDS

322286 Select the option with correct property

1

[Ni (CO)_{4}]

diamagnetic,

{[{NiCl}_{4}]}^{2 -}

paramagnetic

2

[Ni (CO)_{4}]

and

{[{NiCl}_{4}]}^{2 -}

both diamagnetic

3

[Ni (CO)_{4}]

and

{[{NiCl}_{4}]}^{2 -}

both paramagnetic

4

{[{NiCl}_{4}]}^{2 -}

diamagnetic,

[Ni (CO)_{4}]

paramagnetic

Explanation:

${[{NiCl}_{4}]}^{2 -}$ is paramagnetic since it contains two unpaired electrons ( ${sp}^{3}$ hybridization).
$[Ni (CO)_{4}]$ has tetrahedral geometry but it is diamagnetic because nickel is in zero oxidation state and contains no unpaired electron ( ${sp}^{3}$ hybridization). So, the correct option is (1).

JEE Main - 2024

CHXII09:COORDINATION COMPOUNDS

322287 Which of the following is true about the complex $[{PtCl}_{2} ({NH}_{3}) ({OH}_{2})]$ ? [Atomic number of $Pt =$ 78]
(i) It will have two geometrical isomeric forms, cis and trans
(ii) The hybridisation state of $Pt (II)$ is ${sp}^{3}$
(iii) It is a square planar complex
(iv) It is a diamagnetic complex
(v) It can show hydrate isomerism
(vi) It is a tetrahedral complex

1 (i), (iii), (iv)

2 (ii), (iv), (v)

3 (ii), (v), (vi)

4 (i), (v), (vi)

Explanation:

$[{PtCl}_{2} ({NH}_{3}) (H_{2} O)]$ is a square planar molecule with no unpaired electrons. So, it is diamagnetic. It is of ${Ma}_{2} bc$ type molecule. So exhibits geometrical isomerism.

CHXII09:COORDINATION COMPOUNDS

322288 A magnetic moment of 1.73 B.M. will be shown by one among the following:

1

{[{CoCl}_{6}]}^{4 -}

2

{[Cu {({NH}_{3})}_{4}]}^{2 +}

3

{[Ni (CN)_{4}]}^{2 -}

4

{TiCl}_{4}

Explanation:

$∵ \sqrt{n(n + 2)} = 1.73$
$∵ n = 1,$ i.e., species has one unpaired electron.
Thus species is ${[Cu {({NH}_{3})}_{4}]}^{2 +} {Cu}^{2 +}$ in
${[Cu {({NH}_{3})}_{4}]}^{2 +}$ has configuration
supporting img
One electron is shifted from $3d to 4p$ in complex formation.

CHXII09:COORDINATION COMPOUNDS

322289 Among the following complexes ( $K$ to $P$ ), $K_{3} [Fe (CN)_{6}] (K), [Co {({NH}_{3})}_{6}] {Cl}_{3} (L)$ , ${Na}_{3} [Co (oxalate)_{3}] (M), [Ni {(H_{2} O)}_{6}] {Cl}_{2} (N)$ , $K_{2} [Pt (CN)_{4}] (O)$ , and $[Zn {(H_{2} O)}_{6}] {({NO}_{3})}_{2} (P)$ , the diamagnetic complexes are:

1 K, L, M, N

2 K, M, O, P

3 L, M, O, P

4 L, M, N, O

Explanation:

$[K] : K_{3} [Fe(CN)_{6}] : {Fe}^{3 +}$ has $3 d^{5}$ configuration; ${CN}^{-}$ is strong field ligand and thus four electrons are paired leaving one unpaired electron with $d^{2} s p^{3}$ hybridisation and thus paramagnetic.
[L]: $[Co {({NH}_{3})}_{6}] {Cl}_{3} : {Co}^{3 +}$ has configuration; ${NH}_{3}$ is strong field ligand thus all six electrons are paired with $d^{2} s p^{3}$ hybridisation and therefore diamagnetic.
$[M] : {Na}_{3} [Co (ox)_{3}] : {Co}^{3 +}$ configuration; $C_{2} O_{4}^{2 -}$ is strong field ligand and thus all the six electrons are paired with $d^{2} s p^{3}$ hybridisation and therefore diamagnetic.
[N]: $[Ni {(H_{2} O)}_{6}] {Cl}_{2} : {Ni}^{2 +}$ has $3 d^{8}$ configuration with two unpaired electrons. $H_{2} O$ is weak field ligand and thus $s p^{3} d^{2}$ hybridisation and paramagnetic.
[O]: $K_{2} [Pt(CN)_{4}]$ . $; P^{2 +}$ has $3 d^{8}$ Configuration. $C N^{-}$ is strong ligand and thus all the eight electrons are paired with $ds p^{3}$ hybridiastion and therefore diamagnetic .
$[P]: [Zn {(H_{2} O)}_{6} {(N O_{3})}_{2} :Z n^{2 +}$ has $3 d^{10}$ Configuration and thus all paired showing $s p^{3} d^{2}$ hubridisation and so diamagnetic.

CHXII09:COORDINATION COMPOUNDS

322285 Which one of the following is an inner orbital complex as well as diamagnetic in behaviour? (Atomic number: $Zn = 30, Cr = 24, Co = 27, Ni$ $= 28$ )

1

{[Zn {({NH}_{3})}_{6}]}^{2 +}

2

{[Cr {({NH}_{3})}_{6}]}^{3 +}

3

{[Co {({NH}_{3})}_{6}]}^{3 +}

4

{[Ni {({NH}_{3})}_{6}]}^{2 +}

Explanation:

${[Co {({NH}_{3})}_{6}]}^{3 +}, {Co}^{3 +} - [Ar] 3 d^{6}$
supporting img
${[Cr {({NH}_{3})}_{6}]}^{3 +}, {Cr}^{3 +} - [Ar] 3 d^{3}$

${[Ni {({NH}_{3})}_{6}]}^{2 +}, {Ni}^{2 +} - [Ar] 3 d^{8}$

${Zn {({NH}_{3})}_{6}]}^{2 +}, {Zn}^{2 +} - [Ar] 3 d^{10}$

CHXII09:COORDINATION COMPOUNDS

322286 Select the option with correct property

1

[Ni (CO)_{4}]

diamagnetic,

{[{NiCl}_{4}]}^{2 -}

paramagnetic

2

[Ni (CO)_{4}]

and

{[{NiCl}_{4}]}^{2 -}

both diamagnetic

3

[Ni (CO)_{4}]

and

{[{NiCl}_{4}]}^{2 -}

both paramagnetic

4

{[{NiCl}_{4}]}^{2 -}

diamagnetic,

[Ni (CO)_{4}]

paramagnetic

Explanation:

${[{NiCl}_{4}]}^{2 -}$ is paramagnetic since it contains two unpaired electrons ( ${sp}^{3}$ hybridization).
$[Ni (CO)_{4}]$ has tetrahedral geometry but it is diamagnetic because nickel is in zero oxidation state and contains no unpaired electron ( ${sp}^{3}$ hybridization). So, the correct option is (1).

JEE Main - 2024

CHXII09:COORDINATION COMPOUNDS

322287 Which of the following is true about the complex $[{PtCl}_{2} ({NH}_{3}) ({OH}_{2})]$ ? [Atomic number of $Pt =$ 78]
(i) It will have two geometrical isomeric forms, cis and trans
(ii) The hybridisation state of $Pt (II)$ is ${sp}^{3}$
(iii) It is a square planar complex
(iv) It is a diamagnetic complex
(v) It can show hydrate isomerism
(vi) It is a tetrahedral complex

1 (i), (iii), (iv)

2 (ii), (iv), (v)

3 (ii), (v), (vi)

4 (i), (v), (vi)

Explanation:

$[{PtCl}_{2} ({NH}_{3}) (H_{2} O)]$ is a square planar molecule with no unpaired electrons. So, it is diamagnetic. It is of ${Ma}_{2} bc$ type molecule. So exhibits geometrical isomerism.

CHXII09:COORDINATION COMPOUNDS

322288 A magnetic moment of 1.73 B.M. will be shown by one among the following:

1

{[{CoCl}_{6}]}^{4 -}

2

{[Cu {({NH}_{3})}_{4}]}^{2 +}

3

{[Ni (CN)_{4}]}^{2 -}

4

{TiCl}_{4}

Explanation:

$∵ \sqrt{n(n + 2)} = 1.73$
$∵ n = 1,$ i.e., species has one unpaired electron.
Thus species is ${[Cu {({NH}_{3})}_{4}]}^{2 +} {Cu}^{2 +}$ in
${[Cu {({NH}_{3})}_{4}]}^{2 +}$ has configuration
supporting img
One electron is shifted from $3d to 4p$ in complex formation.

CHXII09:COORDINATION COMPOUNDS

322289 Among the following complexes ( $K$ to $P$ ), $K_{3} [Fe (CN)_{6}] (K), [Co {({NH}_{3})}_{6}] {Cl}_{3} (L)$ , ${Na}_{3} [Co (oxalate)_{3}] (M), [Ni {(H_{2} O)}_{6}] {Cl}_{2} (N)$ , $K_{2} [Pt (CN)_{4}] (O)$ , and $[Zn {(H_{2} O)}_{6}] {({NO}_{3})}_{2} (P)$ , the diamagnetic complexes are:

1 K, L, M, N

2 K, M, O, P

3 L, M, O, P

4 L, M, N, O

Explanation:

$[K] : K_{3} [Fe(CN)_{6}] : {Fe}^{3 +}$ has $3 d^{5}$ configuration; ${CN}^{-}$ is strong field ligand and thus four electrons are paired leaving one unpaired electron with $d^{2} s p^{3}$ hybridisation and thus paramagnetic.
[L]: $[Co {({NH}_{3})}_{6}] {Cl}_{3} : {Co}^{3 +}$ has configuration; ${NH}_{3}$ is strong field ligand thus all six electrons are paired with $d^{2} s p^{3}$ hybridisation and therefore diamagnetic.
$[M] : {Na}_{3} [Co (ox)_{3}] : {Co}^{3 +}$ configuration; $C_{2} O_{4}^{2 -}$ is strong field ligand and thus all the six electrons are paired with $d^{2} s p^{3}$ hybridisation and therefore diamagnetic.
[N]: $[Ni {(H_{2} O)}_{6}] {Cl}_{2} : {Ni}^{2 +}$ has $3 d^{8}$ configuration with two unpaired electrons. $H_{2} O$ is weak field ligand and thus $s p^{3} d^{2}$ hybridisation and paramagnetic.
[O]: $K_{2} [Pt(CN)_{4}]$ . $; P^{2 +}$ has $3 d^{8}$ Configuration. $C N^{-}$ is strong ligand and thus all the eight electrons are paired with $ds p^{3}$ hybridiastion and therefore diamagnetic .
$[P]: [Zn {(H_{2} O)}_{6} {(N O_{3})}_{2} :Z n^{2 +}$ has $3 d^{10}$ Configuration and thus all paired showing $s p^{3} d^{2}$ hubridisation and so diamagnetic.

CHXII09:COORDINATION COMPOUNDS

322285 Which one of the following is an inner orbital complex as well as diamagnetic in behaviour? (Atomic number: $Zn = 30, Cr = 24, Co = 27, Ni$ $= 28$ )

1

{[Zn {({NH}_{3})}_{6}]}^{2 +}

2

{[Cr {({NH}_{3})}_{6}]}^{3 +}

3

{[Co {({NH}_{3})}_{6}]}^{3 +}

4

{[Ni {({NH}_{3})}_{6}]}^{2 +}

Explanation:

${[Co {({NH}_{3})}_{6}]}^{3 +}, {Co}^{3 +} - [Ar] 3 d^{6}$
supporting img
${[Cr {({NH}_{3})}_{6}]}^{3 +}, {Cr}^{3 +} - [Ar] 3 d^{3}$

${[Ni {({NH}_{3})}_{6}]}^{2 +}, {Ni}^{2 +} - [Ar] 3 d^{8}$

${Zn {({NH}_{3})}_{6}]}^{2 +}, {Zn}^{2 +} - [Ar] 3 d^{10}$

CHXII09:COORDINATION COMPOUNDS

322286 Select the option with correct property

1

[Ni (CO)_{4}]

diamagnetic,

{[{NiCl}_{4}]}^{2 -}

paramagnetic

2

[Ni (CO)_{4}]

and

{[{NiCl}_{4}]}^{2 -}

both diamagnetic

3

[Ni (CO)_{4}]

and

{[{NiCl}_{4}]}^{2 -}

both paramagnetic

4

{[{NiCl}_{4}]}^{2 -}

diamagnetic,

[Ni (CO)_{4}]

paramagnetic

Explanation:

${[{NiCl}_{4}]}^{2 -}$ is paramagnetic since it contains two unpaired electrons ( ${sp}^{3}$ hybridization).
$[Ni (CO)_{4}]$ has tetrahedral geometry but it is diamagnetic because nickel is in zero oxidation state and contains no unpaired electron ( ${sp}^{3}$ hybridization). So, the correct option is (1).

JEE Main - 2024

CHXII09:COORDINATION COMPOUNDS

322287 Which of the following is true about the complex $[{PtCl}_{2} ({NH}_{3}) ({OH}_{2})]$ ? [Atomic number of $Pt =$ 78]
(i) It will have two geometrical isomeric forms, cis and trans
(ii) The hybridisation state of $Pt (II)$ is ${sp}^{3}$
(iii) It is a square planar complex
(iv) It is a diamagnetic complex
(v) It can show hydrate isomerism
(vi) It is a tetrahedral complex

1 (i), (iii), (iv)

2 (ii), (iv), (v)

3 (ii), (v), (vi)

4 (i), (v), (vi)

Explanation:

$[{PtCl}_{2} ({NH}_{3}) (H_{2} O)]$ is a square planar molecule with no unpaired electrons. So, it is diamagnetic. It is of ${Ma}_{2} bc$ type molecule. So exhibits geometrical isomerism.

CHXII09:COORDINATION COMPOUNDS

322288 A magnetic moment of 1.73 B.M. will be shown by one among the following:

1

{[{CoCl}_{6}]}^{4 -}

2

{[Cu {({NH}_{3})}_{4}]}^{2 +}

3

{[Ni (CN)_{4}]}^{2 -}

4

{TiCl}_{4}

Explanation:

$∵ \sqrt{n(n + 2)} = 1.73$
$∵ n = 1,$ i.e., species has one unpaired electron.
Thus species is ${[Cu {({NH}_{3})}_{4}]}^{2 +} {Cu}^{2 +}$ in
${[Cu {({NH}_{3})}_{4}]}^{2 +}$ has configuration
supporting img
One electron is shifted from $3d to 4p$ in complex formation.

CHXII09:COORDINATION COMPOUNDS

322289 Among the following complexes ( $K$ to $P$ ), $K_{3} [Fe (CN)_{6}] (K), [Co {({NH}_{3})}_{6}] {Cl}_{3} (L)$ , ${Na}_{3} [Co (oxalate)_{3}] (M), [Ni {(H_{2} O)}_{6}] {Cl}_{2} (N)$ , $K_{2} [Pt (CN)_{4}] (O)$ , and $[Zn {(H_{2} O)}_{6}] {({NO}_{3})}_{2} (P)$ , the diamagnetic complexes are:

1 K, L, M, N

2 K, M, O, P

3 L, M, O, P

4 L, M, N, O

Explanation:

$[K] : K_{3} [Fe(CN)_{6}] : {Fe}^{3 +}$ has $3 d^{5}$ configuration; ${CN}^{-}$ is strong field ligand and thus four electrons are paired leaving one unpaired electron with $d^{2} s p^{3}$ hybridisation and thus paramagnetic.
[L]: $[Co {({NH}_{3})}_{6}] {Cl}_{3} : {Co}^{3 +}$ has configuration; ${NH}_{3}$ is strong field ligand thus all six electrons are paired with $d^{2} s p^{3}$ hybridisation and therefore diamagnetic.
$[M] : {Na}_{3} [Co (ox)_{3}] : {Co}^{3 +}$ configuration; $C_{2} O_{4}^{2 -}$ is strong field ligand and thus all the six electrons are paired with $d^{2} s p^{3}$ hybridisation and therefore diamagnetic.
[N]: $[Ni {(H_{2} O)}_{6}] {Cl}_{2} : {Ni}^{2 +}$ has $3 d^{8}$ configuration with two unpaired electrons. $H_{2} O$ is weak field ligand and thus $s p^{3} d^{2}$ hybridisation and paramagnetic.
[O]: $K_{2} [Pt(CN)_{4}]$ . $; P^{2 +}$ has $3 d^{8}$ Configuration. $C N^{-}$ is strong ligand and thus all the eight electrons are paired with $ds p^{3}$ hybridiastion and therefore diamagnetic .
$[P]: [Zn {(H_{2} O)}_{6} {(N O_{3})}_{2} :Z n^{2 +}$ has $3 d^{10}$ Configuration and thus all paired showing $s p^{3} d^{2}$ hubridisation and so diamagnetic.

CHXII09:COORDINATION COMPOUNDS

322285 Which one of the following is an inner orbital complex as well as diamagnetic in behaviour? (Atomic number: $Zn = 30, Cr = 24, Co = 27, Ni$ $= 28$ )

1

{[Zn {({NH}_{3})}_{6}]}^{2 +}

2

{[Cr {({NH}_{3})}_{6}]}^{3 +}

3

{[Co {({NH}_{3})}_{6}]}^{3 +}

4

{[Ni {({NH}_{3})}_{6}]}^{2 +}

Explanation:

${[Co {({NH}_{3})}_{6}]}^{3 +}, {Co}^{3 +} - [Ar] 3 d^{6}$
supporting img
${[Cr {({NH}_{3})}_{6}]}^{3 +}, {Cr}^{3 +} - [Ar] 3 d^{3}$

${[Ni {({NH}_{3})}_{6}]}^{2 +}, {Ni}^{2 +} - [Ar] 3 d^{8}$

${Zn {({NH}_{3})}_{6}]}^{2 +}, {Zn}^{2 +} - [Ar] 3 d^{10}$

CHXII09:COORDINATION COMPOUNDS

322286 Select the option with correct property

1

[Ni (CO)_{4}]

diamagnetic,

{[{NiCl}_{4}]}^{2 -}

paramagnetic

2

[Ni (CO)_{4}]

and

{[{NiCl}_{4}]}^{2 -}

both diamagnetic

3

[Ni (CO)_{4}]

and

{[{NiCl}_{4}]}^{2 -}

both paramagnetic

4

{[{NiCl}_{4}]}^{2 -}

diamagnetic,

[Ni (CO)_{4}]

paramagnetic

Explanation:

${[{NiCl}_{4}]}^{2 -}$ is paramagnetic since it contains two unpaired electrons ( ${sp}^{3}$ hybridization).
$[Ni (CO)_{4}]$ has tetrahedral geometry but it is diamagnetic because nickel is in zero oxidation state and contains no unpaired electron ( ${sp}^{3}$ hybridization). So, the correct option is (1).

JEE Main - 2024

CHXII09:COORDINATION COMPOUNDS

322287 Which of the following is true about the complex $[{PtCl}_{2} ({NH}_{3}) ({OH}_{2})]$ ? [Atomic number of $Pt =$ 78]
(i) It will have two geometrical isomeric forms, cis and trans
(ii) The hybridisation state of $Pt (II)$ is ${sp}^{3}$
(iii) It is a square planar complex
(iv) It is a diamagnetic complex
(v) It can show hydrate isomerism
(vi) It is a tetrahedral complex

1 (i), (iii), (iv)

2 (ii), (iv), (v)

3 (ii), (v), (vi)

4 (i), (v), (vi)

Explanation:

$[{PtCl}_{2} ({NH}_{3}) (H_{2} O)]$ is a square planar molecule with no unpaired electrons. So, it is diamagnetic. It is of ${Ma}_{2} bc$ type molecule. So exhibits geometrical isomerism.

CHXII09:COORDINATION COMPOUNDS

322288 A magnetic moment of 1.73 B.M. will be shown by one among the following:

1

{[{CoCl}_{6}]}^{4 -}

2

{[Cu {({NH}_{3})}_{4}]}^{2 +}

3

{[Ni (CN)_{4}]}^{2 -}

4

{TiCl}_{4}

Explanation:

$∵ \sqrt{n(n + 2)} = 1.73$
$∵ n = 1,$ i.e., species has one unpaired electron.
Thus species is ${[Cu {({NH}_{3})}_{4}]}^{2 +} {Cu}^{2 +}$ in
${[Cu {({NH}_{3})}_{4}]}^{2 +}$ has configuration
supporting img
One electron is shifted from $3d to 4p$ in complex formation.

CHXII09:COORDINATION COMPOUNDS

322289 Among the following complexes ( $K$ to $P$ ), $K_{3} [Fe (CN)_{6}] (K), [Co {({NH}_{3})}_{6}] {Cl}_{3} (L)$ , ${Na}_{3} [Co (oxalate)_{3}] (M), [Ni {(H_{2} O)}_{6}] {Cl}_{2} (N)$ , $K_{2} [Pt (CN)_{4}] (O)$ , and $[Zn {(H_{2} O)}_{6}] {({NO}_{3})}_{2} (P)$ , the diamagnetic complexes are:

1 K, L, M, N

2 K, M, O, P

3 L, M, O, P

4 L, M, N, O

Explanation:

$[K] : K_{3} [Fe(CN)_{6}] : {Fe}^{3 +}$ has $3 d^{5}$ configuration; ${CN}^{-}$ is strong field ligand and thus four electrons are paired leaving one unpaired electron with $d^{2} s p^{3}$ hybridisation and thus paramagnetic.
[L]: $[Co {({NH}_{3})}_{6}] {Cl}_{3} : {Co}^{3 +}$ has configuration; ${NH}_{3}$ is strong field ligand thus all six electrons are paired with $d^{2} s p^{3}$ hybridisation and therefore diamagnetic.
$[M] : {Na}_{3} [Co (ox)_{3}] : {Co}^{3 +}$ configuration; $C_{2} O_{4}^{2 -}$ is strong field ligand and thus all the six electrons are paired with $d^{2} s p^{3}$ hybridisation and therefore diamagnetic.
[N]: $[Ni {(H_{2} O)}_{6}] {Cl}_{2} : {Ni}^{2 +}$ has $3 d^{8}$ configuration with two unpaired electrons. $H_{2} O$ is weak field ligand and thus $s p^{3} d^{2}$ hybridisation and paramagnetic.
[O]: $K_{2} [Pt(CN)_{4}]$ . $; P^{2 +}$ has $3 d^{8}$ Configuration. $C N^{-}$ is strong ligand and thus all the eight electrons are paired with $ds p^{3}$ hybridiastion and therefore diamagnetic .
$[P]: [Zn {(H_{2} O)}_{6} {(N O_{3})}_{2} :Z n^{2 +}$ has $3 d^{10}$ Configuration and thus all paired showing $s p^{3} d^{2}$ hubridisation and so diamagnetic.

CHXII09:COORDINATION COMPOUNDS

322285 Which one of the following is an inner orbital complex as well as diamagnetic in behaviour? (Atomic number: $Zn = 30, Cr = 24, Co = 27, Ni$ $= 28$ )

1

{[Zn {({NH}_{3})}_{6}]}^{2 +}

2

{[Cr {({NH}_{3})}_{6}]}^{3 +}

3

{[Co {({NH}_{3})}_{6}]}^{3 +}

4

{[Ni {({NH}_{3})}_{6}]}^{2 +}

Explanation:

${[Co {({NH}_{3})}_{6}]}^{3 +}, {Co}^{3 +} - [Ar] 3 d^{6}$
supporting img
${[Cr {({NH}_{3})}_{6}]}^{3 +}, {Cr}^{3 +} - [Ar] 3 d^{3}$

${[Ni {({NH}_{3})}_{6}]}^{2 +}, {Ni}^{2 +} - [Ar] 3 d^{8}$

${Zn {({NH}_{3})}_{6}]}^{2 +}, {Zn}^{2 +} - [Ar] 3 d^{10}$

CHXII09:COORDINATION COMPOUNDS

322286 Select the option with correct property

1

[Ni (CO)_{4}]

diamagnetic,

{[{NiCl}_{4}]}^{2 -}

paramagnetic

2

[Ni (CO)_{4}]

and

{[{NiCl}_{4}]}^{2 -}

both diamagnetic

3

[Ni (CO)_{4}]

and

{[{NiCl}_{4}]}^{2 -}

both paramagnetic

4

{[{NiCl}_{4}]}^{2 -}

diamagnetic,

[Ni (CO)_{4}]

paramagnetic

Explanation:

${[{NiCl}_{4}]}^{2 -}$ is paramagnetic since it contains two unpaired electrons ( ${sp}^{3}$ hybridization).
$[Ni (CO)_{4}]$ has tetrahedral geometry but it is diamagnetic because nickel is in zero oxidation state and contains no unpaired electron ( ${sp}^{3}$ hybridization). So, the correct option is (1).

JEE Main - 2024

CHXII09:COORDINATION COMPOUNDS

322287 Which of the following is true about the complex $[{PtCl}_{2} ({NH}_{3}) ({OH}_{2})]$ ? [Atomic number of $Pt =$ 78]
(i) It will have two geometrical isomeric forms, cis and trans
(ii) The hybridisation state of $Pt (II)$ is ${sp}^{3}$
(iii) It is a square planar complex
(iv) It is a diamagnetic complex
(v) It can show hydrate isomerism
(vi) It is a tetrahedral complex

1 (i), (iii), (iv)

2 (ii), (iv), (v)

3 (ii), (v), (vi)

4 (i), (v), (vi)

Explanation:

$[{PtCl}_{2} ({NH}_{3}) (H_{2} O)]$ is a square planar molecule with no unpaired electrons. So, it is diamagnetic. It is of ${Ma}_{2} bc$ type molecule. So exhibits geometrical isomerism.

CHXII09:COORDINATION COMPOUNDS

322288 A magnetic moment of 1.73 B.M. will be shown by one among the following:

1

{[{CoCl}_{6}]}^{4 -}

2

{[Cu {({NH}_{3})}_{4}]}^{2 +}

3

{[Ni (CN)_{4}]}^{2 -}

4

{TiCl}_{4}

Explanation:

$∵ \sqrt{n(n + 2)} = 1.73$
$∵ n = 1,$ i.e., species has one unpaired electron.
Thus species is ${[Cu {({NH}_{3})}_{4}]}^{2 +} {Cu}^{2 +}$ in
${[Cu {({NH}_{3})}_{4}]}^{2 +}$ has configuration
supporting img
One electron is shifted from $3d to 4p$ in complex formation.

CHXII09:COORDINATION COMPOUNDS

322289 Among the following complexes ( $K$ to $P$ ), $K_{3} [Fe (CN)_{6}] (K), [Co {({NH}_{3})}_{6}] {Cl}_{3} (L)$ , ${Na}_{3} [Co (oxalate)_{3}] (M), [Ni {(H_{2} O)}_{6}] {Cl}_{2} (N)$ , $K_{2} [Pt (CN)_{4}] (O)$ , and $[Zn {(H_{2} O)}_{6}] {({NO}_{3})}_{2} (P)$ , the diamagnetic complexes are:

1 K, L, M, N

2 K, M, O, P

3 L, M, O, P

4 L, M, N, O

Explanation:

$[K] : K_{3} [Fe(CN)_{6}] : {Fe}^{3 +}$ has $3 d^{5}$ configuration; ${CN}^{-}$ is strong field ligand and thus four electrons are paired leaving one unpaired electron with $d^{2} s p^{3}$ hybridisation and thus paramagnetic.
[L]: $[Co {({NH}_{3})}_{6}] {Cl}_{3} : {Co}^{3 +}$ has configuration; ${NH}_{3}$ is strong field ligand thus all six electrons are paired with $d^{2} s p^{3}$ hybridisation and therefore diamagnetic.
$[M] : {Na}_{3} [Co (ox)_{3}] : {Co}^{3 +}$ configuration; $C_{2} O_{4}^{2 -}$ is strong field ligand and thus all the six electrons are paired with $d^{2} s p^{3}$ hybridisation and therefore diamagnetic.
[N]: $[Ni {(H_{2} O)}_{6}] {Cl}_{2} : {Ni}^{2 +}$ has $3 d^{8}$ configuration with two unpaired electrons. $H_{2} O$ is weak field ligand and thus $s p^{3} d^{2}$ hybridisation and paramagnetic.
[O]: $K_{2} [Pt(CN)_{4}]$ . $; P^{2 +}$ has $3 d^{8}$ Configuration. $C N^{-}$ is strong ligand and thus all the eight electrons are paired with $ds p^{3}$ hybridiastion and therefore diamagnetic .
$[P]: [Zn {(H_{2} O)}_{6} {(N O_{3})}_{2} :Z n^{2 +}$ has $3 d^{10}$ Configuration and thus all paired showing $s p^{3} d^{2}$ hubridisation and so diamagnetic.