QBManager

COORDINATION COMPOUNDS

274149 Which of the following is the correct configuration of the complex ${[{RhCl}_{6}]}^{3 -}$ ?

1 High spin

t_{2 g}^{4} e_{g}_{g}

2 Low spin

t_{2 g}^{6} e_{g}

3 Low spin

t_{2 g}^{3} e_{g}_{g}^{g}

4 High spin

t_{2 g}^{5} e_{g}_{g}

Explanation:

(B) : ${[{RhCl}_{6}]}^{3 -}$
Let, $x$ be the oxidation state of $Rh$ .
$x + 6 (- 1) = - 3$
$x = + 3$
$Rh = [Kr] 4 d^{7} 5 s^{2}$
${Rh}^{3 +} = [Kr] 4 d^{6} (∵$ Pairing occur because $Rh$ belongs to $4 d$ series)
original image
It forms the low spin complex because electron fill with forcefully in $t_{2 g}$ orbital i.e. low $spin t t_{2 g}^{6} e_{g}^{0}$

J and K CET-(2018)

COORDINATION COMPOUNDS

274150 The number of d-electrons in ${[Cr {(H_{2} O)}_{6}]}^{3 +}$ is

1 2

2 3

3 4

4 5

Explanation:

(B) : ${[Cr {(H_{2} O)}_{6}]}^{3 +}$
The oxidation state of $Cr$ is +3 .
$Cr = [Ar] 4 s^{1} 3 d^{5}$
${Cr}^{+ 3} = [Ar] 4 s^{0} 3 d^{3}$
$∴$ The d-electron in ${[Cr {(H_{2} O)}_{4}]}^{3 +}$ is 3 .

MPPET- 2009

COORDINATION COMPOUNDS

274151 The hybridization of $Fe$ in $K_{3} Fe (CN)_{6}$ is

1

{sp}^{3}

2

{dsp}^{3}

3

{sp}^{3} d^{2}

4

d^{2} {sp}^{3}

Explanation:

(D) : $K_{3} [Fe (CN)_{6}]$
Let, $x$ be the oxidation state of $Fe$ .
$3 (+ 1) + x + 6 (- 1) = 0$
$x = + 3$
${Fe}^{3 +} = [Ar] 3 d^{5} (∵$ Pairing occur because $CN$ is $SFL)$
original image
Hence, the hybridization of $K_{3} [Fe (CN)_{6}]$ is $d^{2} {sp}^{3}$ .

BITSAT-2016

COORDINATION COMPOUNDS

274158 In ${[{NiCl}_{4}]}^{2 -}$ , the type of hybridization for $Ni$ is

1

{sp}^{3} d^{2}

2

{dsp}^{3}

3

{sp}^{3}

4

{dsp}^{2}

Explanation:

(C) : For the complex-
${[{NiCl}_{4}]}^{2 -}$
Let, $x$ be the oxidation state of $Ni$
$∴ x + 4 (- 1) = - 2$
$x = + 2$
${Ni}^{2 +} = [Ar] 3 d^{8}$
$Cl$ is a weak field ligand thus the pairing is not occur.
original image
Hence, the hybridization of ${[Ni (Cl)_{4}]}^{2 -}$ is ${sp}^{3}$ .

AMU-2012

COORDINATION COMPOUNDS

274149 Which of the following is the correct configuration of the complex ${[{RhCl}_{6}]}^{3 -}$ ?

1 High spin

t_{2 g}^{4} e_{g}_{g}

2 Low spin

t_{2 g}^{6} e_{g}

3 Low spin

t_{2 g}^{3} e_{g}_{g}^{g}

4 High spin

t_{2 g}^{5} e_{g}_{g}

Explanation:

(B) : ${[{RhCl}_{6}]}^{3 -}$
Let, $x$ be the oxidation state of $Rh$ .
$x + 6 (- 1) = - 3$
$x = + 3$
$Rh = [Kr] 4 d^{7} 5 s^{2}$
${Rh}^{3 +} = [Kr] 4 d^{6} (∵$ Pairing occur because $Rh$ belongs to $4 d$ series)
original image
It forms the low spin complex because electron fill with forcefully in $t_{2 g}$ orbital i.e. low $spin t t_{2 g}^{6} e_{g}^{0}$

J and K CET-(2018)

COORDINATION COMPOUNDS

274150 The number of d-electrons in ${[Cr {(H_{2} O)}_{6}]}^{3 +}$ is

1 2

2 3

3 4

4 5

Explanation:

(B) : ${[Cr {(H_{2} O)}_{6}]}^{3 +}$
The oxidation state of $Cr$ is +3 .
$Cr = [Ar] 4 s^{1} 3 d^{5}$
${Cr}^{+ 3} = [Ar] 4 s^{0} 3 d^{3}$
$∴$ The d-electron in ${[Cr {(H_{2} O)}_{4}]}^{3 +}$ is 3 .

MPPET- 2009

COORDINATION COMPOUNDS

274151 The hybridization of $Fe$ in $K_{3} Fe (CN)_{6}$ is

1

{sp}^{3}

2

{dsp}^{3}

3

{sp}^{3} d^{2}

4

d^{2} {sp}^{3}

Explanation:

(D) : $K_{3} [Fe (CN)_{6}]$
Let, $x$ be the oxidation state of $Fe$ .
$3 (+ 1) + x + 6 (- 1) = 0$
$x = + 3$
${Fe}^{3 +} = [Ar] 3 d^{5} (∵$ Pairing occur because $CN$ is $SFL)$
original image
Hence, the hybridization of $K_{3} [Fe (CN)_{6}]$ is $d^{2} {sp}^{3}$ .

BITSAT-2016

COORDINATION COMPOUNDS

274158 In ${[{NiCl}_{4}]}^{2 -}$ , the type of hybridization for $Ni$ is

1

{sp}^{3} d^{2}

2

{dsp}^{3}

3

{sp}^{3}

4

{dsp}^{2}

Explanation:

(C) : For the complex-
${[{NiCl}_{4}]}^{2 -}$
Let, $x$ be the oxidation state of $Ni$
$∴ x + 4 (- 1) = - 2$
$x = + 2$
${Ni}^{2 +} = [Ar] 3 d^{8}$
$Cl$ is a weak field ligand thus the pairing is not occur.
original image
Hence, the hybridization of ${[Ni (Cl)_{4}]}^{2 -}$ is ${sp}^{3}$ .

AMU-2012

COORDINATION COMPOUNDS

274149 Which of the following is the correct configuration of the complex ${[{RhCl}_{6}]}^{3 -}$ ?

1 High spin

t_{2 g}^{4} e_{g}_{g}

2 Low spin

t_{2 g}^{6} e_{g}

3 Low spin

t_{2 g}^{3} e_{g}_{g}^{g}

4 High spin

t_{2 g}^{5} e_{g}_{g}

Explanation:

(B) : ${[{RhCl}_{6}]}^{3 -}$
Let, $x$ be the oxidation state of $Rh$ .
$x + 6 (- 1) = - 3$
$x = + 3$
$Rh = [Kr] 4 d^{7} 5 s^{2}$
${Rh}^{3 +} = [Kr] 4 d^{6} (∵$ Pairing occur because $Rh$ belongs to $4 d$ series)
original image
It forms the low spin complex because electron fill with forcefully in $t_{2 g}$ orbital i.e. low $spin t t_{2 g}^{6} e_{g}^{0}$

J and K CET-(2018)

COORDINATION COMPOUNDS

274150 The number of d-electrons in ${[Cr {(H_{2} O)}_{6}]}^{3 +}$ is

1 2

2 3

3 4

4 5

Explanation:

(B) : ${[Cr {(H_{2} O)}_{6}]}^{3 +}$
The oxidation state of $Cr$ is +3 .
$Cr = [Ar] 4 s^{1} 3 d^{5}$
${Cr}^{+ 3} = [Ar] 4 s^{0} 3 d^{3}$
$∴$ The d-electron in ${[Cr {(H_{2} O)}_{4}]}^{3 +}$ is 3 .

MPPET- 2009

COORDINATION COMPOUNDS

274151 The hybridization of $Fe$ in $K_{3} Fe (CN)_{6}$ is

1

{sp}^{3}

2

{dsp}^{3}

3

{sp}^{3} d^{2}

4

d^{2} {sp}^{3}

Explanation:

(D) : $K_{3} [Fe (CN)_{6}]$
Let, $x$ be the oxidation state of $Fe$ .
$3 (+ 1) + x + 6 (- 1) = 0$
$x = + 3$
${Fe}^{3 +} = [Ar] 3 d^{5} (∵$ Pairing occur because $CN$ is $SFL)$
original image
Hence, the hybridization of $K_{3} [Fe (CN)_{6}]$ is $d^{2} {sp}^{3}$ .

BITSAT-2016

COORDINATION COMPOUNDS

274158 In ${[{NiCl}_{4}]}^{2 -}$ , the type of hybridization for $Ni$ is

1

{sp}^{3} d^{2}

2

{dsp}^{3}

3

{sp}^{3}

4

{dsp}^{2}

Explanation:

(C) : For the complex-
${[{NiCl}_{4}]}^{2 -}$
Let, $x$ be the oxidation state of $Ni$
$∴ x + 4 (- 1) = - 2$
$x = + 2$
${Ni}^{2 +} = [Ar] 3 d^{8}$
$Cl$ is a weak field ligand thus the pairing is not occur.
original image
Hence, the hybridization of ${[Ni (Cl)_{4}]}^{2 -}$ is ${sp}^{3}$ .

AMU-2012

COORDINATION COMPOUNDS

274149 Which of the following is the correct configuration of the complex ${[{RhCl}_{6}]}^{3 -}$ ?

1 High spin

t_{2 g}^{4} e_{g}_{g}

2 Low spin

t_{2 g}^{6} e_{g}

3 Low spin

t_{2 g}^{3} e_{g}_{g}^{g}

4 High spin

t_{2 g}^{5} e_{g}_{g}

Explanation:

(B) : ${[{RhCl}_{6}]}^{3 -}$
Let, $x$ be the oxidation state of $Rh$ .
$x + 6 (- 1) = - 3$
$x = + 3$
$Rh = [Kr] 4 d^{7} 5 s^{2}$
${Rh}^{3 +} = [Kr] 4 d^{6} (∵$ Pairing occur because $Rh$ belongs to $4 d$ series)
original image
It forms the low spin complex because electron fill with forcefully in $t_{2 g}$ orbital i.e. low $spin t t_{2 g}^{6} e_{g}^{0}$

J and K CET-(2018)

COORDINATION COMPOUNDS

274150 The number of d-electrons in ${[Cr {(H_{2} O)}_{6}]}^{3 +}$ is

1 2

2 3

3 4

4 5

Explanation:

(B) : ${[Cr {(H_{2} O)}_{6}]}^{3 +}$
The oxidation state of $Cr$ is +3 .
$Cr = [Ar] 4 s^{1} 3 d^{5}$
${Cr}^{+ 3} = [Ar] 4 s^{0} 3 d^{3}$
$∴$ The d-electron in ${[Cr {(H_{2} O)}_{4}]}^{3 +}$ is 3 .

MPPET- 2009

COORDINATION COMPOUNDS

274151 The hybridization of $Fe$ in $K_{3} Fe (CN)_{6}$ is

1

{sp}^{3}

2

{dsp}^{3}

3

{sp}^{3} d^{2}

4

d^{2} {sp}^{3}

Explanation:

(D) : $K_{3} [Fe (CN)_{6}]$
Let, $x$ be the oxidation state of $Fe$ .
$3 (+ 1) + x + 6 (- 1) = 0$
$x = + 3$
${Fe}^{3 +} = [Ar] 3 d^{5} (∵$ Pairing occur because $CN$ is $SFL)$
original image
Hence, the hybridization of $K_{3} [Fe (CN)_{6}]$ is $d^{2} {sp}^{3}$ .

BITSAT-2016

COORDINATION COMPOUNDS

274158 In ${[{NiCl}_{4}]}^{2 -}$ , the type of hybridization for $Ni$ is

1

{sp}^{3} d^{2}

2

{dsp}^{3}

3

{sp}^{3}

4

{dsp}^{2}

Explanation:

(C) : For the complex-
${[{NiCl}_{4}]}^{2 -}$
Let, $x$ be the oxidation state of $Ni$
$∴ x + 4 (- 1) = - 2$
$x = + 2$
${Ni}^{2 +} = [Ar] 3 d^{8}$
$Cl$ is a weak field ligand thus the pairing is not occur.
original image
Hence, the hybridization of ${[Ni (Cl)_{4}]}^{2 -}$ is ${sp}^{3}$ .

AMU-2012