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CHXII09:COORDINATION COMPOUNDS

322273 ${[Co {({NH}_{3})}_{6}]}^{3 +}$ and ${[{CoF}_{6}]}^{3 -}$ are respectively known as

1 inner orbital complex, spin paired complex

2 spin paired complex, spin free complex

3 spin free complex, spin paired complex

4 outer orbital complex, inner orbital complex

Explanation:

${NH}_{3}$ is strong ligand in case of ${Co}^{+ 3}$ and in ${[Co {({NH}_{3})}_{6}]}^{+ 3}$ compound, ${Co}^{+ 3}$ undergoes $d^{2} {sp}^{3}$ hybridization. It is an inner orbital or low spin or spin paired complex.
In ${[{CoF}_{6}]}^{3 -}, {Co}^{+ 3}$ uses outer 4 d orbital in hybridisation $({sp}^{3} d^{2})$ . It is thus called outer orbital or high spin or spin free complex. So, the correct option is (2).

JEE Main - 2024

CHXII09:COORDINATION COMPOUNDS

322274 ${[Co {({NH}_{3})}_{6}]}^{3 +}$ is an inner orbital complex whereas ${[Ni {({NH}_{3})}_{6}]}^{2 +}$ is an outer orbital complex because

1

{Co}^{3 +}

has

d^{6}

whereas

{Ni}^{2 +}

has

d^{8}

hence pairing is not possible in

{Ni}^{2 +}

2 Ammonia behaves differently with both the metals

3 The approach of ligands to central metal ion is different.

4 No scientific reason

Explanation:

In the presence of ${NH}_{3}$ , the $3 d$ electrons pair up leaving two d-orbitals empty to be involved in $d^{2} {sp}^{3}$ hybridisation forming inner orbital complex in case of ${[Co {({NH}_{3})}_{6}]}^{3 +}$ .
In $Ni {({NH}_{3})}_{6}_{6} +$ , $Ni$ is in +2 oxidation state and has $d^{8}$ configuration, the hybridisation involved is ${sp}^{3} d^{2}$ forming outer orbital complex.

CHXII09:COORDINATION COMPOUNDS

322275 The number of unpaired d- electrons in ${[Co {(H_{2} O)}_{6}]}^{3 +}$ is ____.

1 4

2 2

3 1

4 0

Explanation:

supporting img
In this complex, $d^{2} {sp}^{3}$ hybridisation takes place by pairing of four unpaired electrons into two pairs.
$∴$ Number of unpaired electrons in the complex is zero. So, the correct option is (4).

JEE Main - 2024

CHXII09:COORDINATION COMPOUNDS

322276 In which of the following there is outer orbital complex

1

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

2

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

3

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

4

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

Explanation:

$In {[Zn {({NH}_{3})}_{6}]}^{2 +}, {sp}^{3} d^{2}$ hybridisation results in an outer orbital complex.

CHXII09:COORDINATION COMPOUNDS

322273 ${[Co {({NH}_{3})}_{6}]}^{3 +}$ and ${[{CoF}_{6}]}^{3 -}$ are respectively known as

1 inner orbital complex, spin paired complex

2 spin paired complex, spin free complex

3 spin free complex, spin paired complex

4 outer orbital complex, inner orbital complex

Explanation:

${NH}_{3}$ is strong ligand in case of ${Co}^{+ 3}$ and in ${[Co {({NH}_{3})}_{6}]}^{+ 3}$ compound, ${Co}^{+ 3}$ undergoes $d^{2} {sp}^{3}$ hybridization. It is an inner orbital or low spin or spin paired complex.
In ${[{CoF}_{6}]}^{3 -}, {Co}^{+ 3}$ uses outer 4 d orbital in hybridisation $({sp}^{3} d^{2})$ . It is thus called outer orbital or high spin or spin free complex. So, the correct option is (2).

JEE Main - 2024

CHXII09:COORDINATION COMPOUNDS

322274 ${[Co {({NH}_{3})}_{6}]}^{3 +}$ is an inner orbital complex whereas ${[Ni {({NH}_{3})}_{6}]}^{2 +}$ is an outer orbital complex because

1

{Co}^{3 +}

has

d^{6}

whereas

{Ni}^{2 +}

has

d^{8}

hence pairing is not possible in

{Ni}^{2 +}

2 Ammonia behaves differently with both the metals

3 The approach of ligands to central metal ion is different.

4 No scientific reason

Explanation:

In the presence of ${NH}_{3}$ , the $3 d$ electrons pair up leaving two d-orbitals empty to be involved in $d^{2} {sp}^{3}$ hybridisation forming inner orbital complex in case of ${[Co {({NH}_{3})}_{6}]}^{3 +}$ .
In $Ni {({NH}_{3})}_{6}_{6} +$ , $Ni$ is in +2 oxidation state and has $d^{8}$ configuration, the hybridisation involved is ${sp}^{3} d^{2}$ forming outer orbital complex.

CHXII09:COORDINATION COMPOUNDS

322275 The number of unpaired d- electrons in ${[Co {(H_{2} O)}_{6}]}^{3 +}$ is ____.

1 4

2 2

3 1

4 0

Explanation:

supporting img
In this complex, $d^{2} {sp}^{3}$ hybridisation takes place by pairing of four unpaired electrons into two pairs.
$∴$ Number of unpaired electrons in the complex is zero. So, the correct option is (4).

JEE Main - 2024

CHXII09:COORDINATION COMPOUNDS

322276 In which of the following there is outer orbital complex

1

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

2

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

3

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

4

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

Explanation:

$In {[Zn {({NH}_{3})}_{6}]}^{2 +}, {sp}^{3} d^{2}$ hybridisation results in an outer orbital complex.

CHXII09:COORDINATION COMPOUNDS

322273 ${[Co {({NH}_{3})}_{6}]}^{3 +}$ and ${[{CoF}_{6}]}^{3 -}$ are respectively known as

1 inner orbital complex, spin paired complex

2 spin paired complex, spin free complex

3 spin free complex, spin paired complex

4 outer orbital complex, inner orbital complex

Explanation:

${NH}_{3}$ is strong ligand in case of ${Co}^{+ 3}$ and in ${[Co {({NH}_{3})}_{6}]}^{+ 3}$ compound, ${Co}^{+ 3}$ undergoes $d^{2} {sp}^{3}$ hybridization. It is an inner orbital or low spin or spin paired complex.
In ${[{CoF}_{6}]}^{3 -}, {Co}^{+ 3}$ uses outer 4 d orbital in hybridisation $({sp}^{3} d^{2})$ . It is thus called outer orbital or high spin or spin free complex. So, the correct option is (2).

JEE Main - 2024

CHXII09:COORDINATION COMPOUNDS

322274 ${[Co {({NH}_{3})}_{6}]}^{3 +}$ is an inner orbital complex whereas ${[Ni {({NH}_{3})}_{6}]}^{2 +}$ is an outer orbital complex because

1

{Co}^{3 +}

has

d^{6}

whereas

{Ni}^{2 +}

has

d^{8}

hence pairing is not possible in

{Ni}^{2 +}

2 Ammonia behaves differently with both the metals

3 The approach of ligands to central metal ion is different.

4 No scientific reason

Explanation:

In the presence of ${NH}_{3}$ , the $3 d$ electrons pair up leaving two d-orbitals empty to be involved in $d^{2} {sp}^{3}$ hybridisation forming inner orbital complex in case of ${[Co {({NH}_{3})}_{6}]}^{3 +}$ .
In $Ni {({NH}_{3})}_{6}_{6} +$ , $Ni$ is in +2 oxidation state and has $d^{8}$ configuration, the hybridisation involved is ${sp}^{3} d^{2}$ forming outer orbital complex.

CHXII09:COORDINATION COMPOUNDS

322275 The number of unpaired d- electrons in ${[Co {(H_{2} O)}_{6}]}^{3 +}$ is ____.

1 4

2 2

3 1

4 0

Explanation:

supporting img
In this complex, $d^{2} {sp}^{3}$ hybridisation takes place by pairing of four unpaired electrons into two pairs.
$∴$ Number of unpaired electrons in the complex is zero. So, the correct option is (4).

JEE Main - 2024

CHXII09:COORDINATION COMPOUNDS

322276 In which of the following there is outer orbital complex

1

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

2

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

3

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

4

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

Explanation:

$In {[Zn {({NH}_{3})}_{6}]}^{2 +}, {sp}^{3} d^{2}$ hybridisation results in an outer orbital complex.

CHXII09:COORDINATION COMPOUNDS

322273 ${[Co {({NH}_{3})}_{6}]}^{3 +}$ and ${[{CoF}_{6}]}^{3 -}$ are respectively known as

1 inner orbital complex, spin paired complex

2 spin paired complex, spin free complex

3 spin free complex, spin paired complex

4 outer orbital complex, inner orbital complex

Explanation:

${NH}_{3}$ is strong ligand in case of ${Co}^{+ 3}$ and in ${[Co {({NH}_{3})}_{6}]}^{+ 3}$ compound, ${Co}^{+ 3}$ undergoes $d^{2} {sp}^{3}$ hybridization. It is an inner orbital or low spin or spin paired complex.
In ${[{CoF}_{6}]}^{3 -}, {Co}^{+ 3}$ uses outer 4 d orbital in hybridisation $({sp}^{3} d^{2})$ . It is thus called outer orbital or high spin or spin free complex. So, the correct option is (2).

JEE Main - 2024

CHXII09:COORDINATION COMPOUNDS

322274 ${[Co {({NH}_{3})}_{6}]}^{3 +}$ is an inner orbital complex whereas ${[Ni {({NH}_{3})}_{6}]}^{2 +}$ is an outer orbital complex because

1

{Co}^{3 +}

has

d^{6}

whereas

{Ni}^{2 +}

has

d^{8}

hence pairing is not possible in

{Ni}^{2 +}

2 Ammonia behaves differently with both the metals

3 The approach of ligands to central metal ion is different.

4 No scientific reason

Explanation:

In the presence of ${NH}_{3}$ , the $3 d$ electrons pair up leaving two d-orbitals empty to be involved in $d^{2} {sp}^{3}$ hybridisation forming inner orbital complex in case of ${[Co {({NH}_{3})}_{6}]}^{3 +}$ .
In $Ni {({NH}_{3})}_{6}_{6} +$ , $Ni$ is in +2 oxidation state and has $d^{8}$ configuration, the hybridisation involved is ${sp}^{3} d^{2}$ forming outer orbital complex.

CHXII09:COORDINATION COMPOUNDS

322275 The number of unpaired d- electrons in ${[Co {(H_{2} O)}_{6}]}^{3 +}$ is ____.

1 4

2 2

3 1

4 0

Explanation:

supporting img
In this complex, $d^{2} {sp}^{3}$ hybridisation takes place by pairing of four unpaired electrons into two pairs.
$∴$ Number of unpaired electrons in the complex is zero. So, the correct option is (4).

JEE Main - 2024

CHXII09:COORDINATION COMPOUNDS

322276 In which of the following there is outer orbital complex

1

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

2

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

3

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

4

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

Explanation:

$In {[Zn {({NH}_{3})}_{6}]}^{2 +}, {sp}^{3} d^{2}$ hybridisation results in an outer orbital complex.

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