322273
\({\mathrm{\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}}}\) and \({\mathrm{\left[\mathrm{CoF}_{6}\right]^{3-}}}\) 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:
\({\mathrm{\mathrm{NH}_{3}}}\) is strong ligand in case of \({\mathrm{\mathrm{Co}^{+3}}}\) and in \({\mathrm{\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{+3}}}\) compound, \({\mathrm{\mathrm{Co}^{+3}}}\) undergoes \({\mathrm{\mathrm{d}^{2} \mathrm{sp}^{3}}}\) hybridization. It is an inner orbital or low spin or spin paired complex. In \({\mathrm{\left[\mathrm{CoF}_{6}\right]^{3-}, \mathrm{Co}^{+3}}}\) uses outer 4 d orbital in hybridisation \({\mathrm{\left(\mathrm{sp}^{3} \mathrm{~d}^{2}\right)}}\). 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
\(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}\) is an inner orbital complex whereas \(\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}\) is an outer orbital complex because
1 \(\mathrm{Co}^{3+}\) has \(\mathrm{d}^{6}\) whereas \(\mathrm{Ni}^{2+}\) has \(\mathrm{d}^{8}\) hence pairing is not possible in \(\mathrm{Ni}^{2+}\)
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 \(\mathrm{NH}_{3}\), the \(3 \mathrm{~d}\) electrons pair up leaving two d-orbitals empty to be involved in \(\mathrm{d}^{2} \mathrm{sp}^{3}\) hybridisation forming inner orbital complex in case of \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}\). In \(\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}{ }_{6}+\), \(\mathrm{Ni}\) is in +2 oxidation state and has \(\mathrm{d}^{8}\) configuration, the hybridisation involved is \(\mathrm{sp}^{3} \mathrm{~d}^{2}\) forming outer orbital complex.
CHXII09:COORDINATION COMPOUNDS
322275
The number of unpaired d- electrons in \({\mathrm{\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}}}\) is ____.
1 4
2 2
3 1
4 0
Explanation:
In this complex, \({\mathrm{\mathrm{d}^{2} \mathrm{sp}^{3}}}\) hybridisation takes place by pairing of four unpaired electrons into two pairs. \({\mathrm{\therefore}}\) 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
\(\operatorname{In}\left[\mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}, \mathrm{sp}^{3} \mathrm{~d}^{2}\) hybridisation results in an outer orbital complex.
322273
\({\mathrm{\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}}}\) and \({\mathrm{\left[\mathrm{CoF}_{6}\right]^{3-}}}\) 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:
\({\mathrm{\mathrm{NH}_{3}}}\) is strong ligand in case of \({\mathrm{\mathrm{Co}^{+3}}}\) and in \({\mathrm{\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{+3}}}\) compound, \({\mathrm{\mathrm{Co}^{+3}}}\) undergoes \({\mathrm{\mathrm{d}^{2} \mathrm{sp}^{3}}}\) hybridization. It is an inner orbital or low spin or spin paired complex. In \({\mathrm{\left[\mathrm{CoF}_{6}\right]^{3-}, \mathrm{Co}^{+3}}}\) uses outer 4 d orbital in hybridisation \({\mathrm{\left(\mathrm{sp}^{3} \mathrm{~d}^{2}\right)}}\). 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
\(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}\) is an inner orbital complex whereas \(\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}\) is an outer orbital complex because
1 \(\mathrm{Co}^{3+}\) has \(\mathrm{d}^{6}\) whereas \(\mathrm{Ni}^{2+}\) has \(\mathrm{d}^{8}\) hence pairing is not possible in \(\mathrm{Ni}^{2+}\)
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 \(\mathrm{NH}_{3}\), the \(3 \mathrm{~d}\) electrons pair up leaving two d-orbitals empty to be involved in \(\mathrm{d}^{2} \mathrm{sp}^{3}\) hybridisation forming inner orbital complex in case of \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}\). In \(\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}{ }_{6}+\), \(\mathrm{Ni}\) is in +2 oxidation state and has \(\mathrm{d}^{8}\) configuration, the hybridisation involved is \(\mathrm{sp}^{3} \mathrm{~d}^{2}\) forming outer orbital complex.
CHXII09:COORDINATION COMPOUNDS
322275
The number of unpaired d- electrons in \({\mathrm{\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}}}\) is ____.
1 4
2 2
3 1
4 0
Explanation:
In this complex, \({\mathrm{\mathrm{d}^{2} \mathrm{sp}^{3}}}\) hybridisation takes place by pairing of four unpaired electrons into two pairs. \({\mathrm{\therefore}}\) 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
\(\operatorname{In}\left[\mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}, \mathrm{sp}^{3} \mathrm{~d}^{2}\) hybridisation results in an outer orbital complex.
322273
\({\mathrm{\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}}}\) and \({\mathrm{\left[\mathrm{CoF}_{6}\right]^{3-}}}\) 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:
\({\mathrm{\mathrm{NH}_{3}}}\) is strong ligand in case of \({\mathrm{\mathrm{Co}^{+3}}}\) and in \({\mathrm{\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{+3}}}\) compound, \({\mathrm{\mathrm{Co}^{+3}}}\) undergoes \({\mathrm{\mathrm{d}^{2} \mathrm{sp}^{3}}}\) hybridization. It is an inner orbital or low spin or spin paired complex. In \({\mathrm{\left[\mathrm{CoF}_{6}\right]^{3-}, \mathrm{Co}^{+3}}}\) uses outer 4 d orbital in hybridisation \({\mathrm{\left(\mathrm{sp}^{3} \mathrm{~d}^{2}\right)}}\). 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
\(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}\) is an inner orbital complex whereas \(\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}\) is an outer orbital complex because
1 \(\mathrm{Co}^{3+}\) has \(\mathrm{d}^{6}\) whereas \(\mathrm{Ni}^{2+}\) has \(\mathrm{d}^{8}\) hence pairing is not possible in \(\mathrm{Ni}^{2+}\)
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 \(\mathrm{NH}_{3}\), the \(3 \mathrm{~d}\) electrons pair up leaving two d-orbitals empty to be involved in \(\mathrm{d}^{2} \mathrm{sp}^{3}\) hybridisation forming inner orbital complex in case of \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}\). In \(\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}{ }_{6}+\), \(\mathrm{Ni}\) is in +2 oxidation state and has \(\mathrm{d}^{8}\) configuration, the hybridisation involved is \(\mathrm{sp}^{3} \mathrm{~d}^{2}\) forming outer orbital complex.
CHXII09:COORDINATION COMPOUNDS
322275
The number of unpaired d- electrons in \({\mathrm{\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}}}\) is ____.
1 4
2 2
3 1
4 0
Explanation:
In this complex, \({\mathrm{\mathrm{d}^{2} \mathrm{sp}^{3}}}\) hybridisation takes place by pairing of four unpaired electrons into two pairs. \({\mathrm{\therefore}}\) 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
\(\operatorname{In}\left[\mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}, \mathrm{sp}^{3} \mathrm{~d}^{2}\) hybridisation results in an outer orbital complex.
322273
\({\mathrm{\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}}}\) and \({\mathrm{\left[\mathrm{CoF}_{6}\right]^{3-}}}\) 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:
\({\mathrm{\mathrm{NH}_{3}}}\) is strong ligand in case of \({\mathrm{\mathrm{Co}^{+3}}}\) and in \({\mathrm{\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{+3}}}\) compound, \({\mathrm{\mathrm{Co}^{+3}}}\) undergoes \({\mathrm{\mathrm{d}^{2} \mathrm{sp}^{3}}}\) hybridization. It is an inner orbital or low spin or spin paired complex. In \({\mathrm{\left[\mathrm{CoF}_{6}\right]^{3-}, \mathrm{Co}^{+3}}}\) uses outer 4 d orbital in hybridisation \({\mathrm{\left(\mathrm{sp}^{3} \mathrm{~d}^{2}\right)}}\). 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
\(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}\) is an inner orbital complex whereas \(\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}\) is an outer orbital complex because
1 \(\mathrm{Co}^{3+}\) has \(\mathrm{d}^{6}\) whereas \(\mathrm{Ni}^{2+}\) has \(\mathrm{d}^{8}\) hence pairing is not possible in \(\mathrm{Ni}^{2+}\)
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 \(\mathrm{NH}_{3}\), the \(3 \mathrm{~d}\) electrons pair up leaving two d-orbitals empty to be involved in \(\mathrm{d}^{2} \mathrm{sp}^{3}\) hybridisation forming inner orbital complex in case of \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}\). In \(\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}{ }_{6}+\), \(\mathrm{Ni}\) is in +2 oxidation state and has \(\mathrm{d}^{8}\) configuration, the hybridisation involved is \(\mathrm{sp}^{3} \mathrm{~d}^{2}\) forming outer orbital complex.
CHXII09:COORDINATION COMPOUNDS
322275
The number of unpaired d- electrons in \({\mathrm{\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}}}\) is ____.
1 4
2 2
3 1
4 0
Explanation:
In this complex, \({\mathrm{\mathrm{d}^{2} \mathrm{sp}^{3}}}\) hybridisation takes place by pairing of four unpaired electrons into two pairs. \({\mathrm{\therefore}}\) 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
\(\operatorname{In}\left[\mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}, \mathrm{sp}^{3} \mathrm{~d}^{2}\) hybridisation results in an outer orbital complex.
