Larger the number of unpaired electrons, the more it has paramagnetism. The configuration of \(\mathrm{Cr}^{3+}\left(3 d^{3}\right), \mathrm{Fe}^{2+}\left(3 d^{6}\right), \mathrm{Cu}^{2+}\left(3 d^{9}\right)\) and \(\mathrm{Zn}^{2+}\left(3 d^{10}\right)\) are outer orbital complex ions. Hence \(\mathrm{Fe}^{2+}\) has maximum unpaired electrons.
CHXII09:COORDINATION COMPOUNDS
322295
The magnetic moment is measured in Bohr Magneton (B.M.). Spin only magnetic moment of Fe in \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\) and \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}\) complexes respectively is
1 5.92 B.M. and 1.732 B.M.
2 3.87 B.M. and 1.732 B.M.
3 4.89 B.M. and 6.92 B.M.
4 6.92 B.M. in both.
Explanation:
In both the complexes, Fe is in \( + \,3\) oxidation state. \(\mathrm{CN}^{-}\) is a strong field ligand, but \(\mathrm{H}_{2} \mathrm{O}\) is a weak field ligand.
There are 5 unpaired electrons. Spin only magnetic moment value \(=\sqrt{\mathrm{n}(\mathrm{n}+2)}=\sqrt{5(5+2)}=\sqrt{35}=5.92\) B.M.
There is only one unpaired electron. Spin only magnetic moment value \(=\sqrt{\mathrm{n}(\mathrm{n}+2)}=\sqrt{1(1+2)}=\sqrt{3}=1.732\) B.M. Thus, option (1) is correct.
JEE - 2023
CHXII09:COORDINATION COMPOUNDS
322296
The \(d\)-electronic configuration of \(\mathrm{Cr}^{2+}, \mathrm{Mn}^{2+}, \mathrm{Fe}^{2+} \mathrm{Ni}^{2+}\) are \(3 d^{4}, 3 d^{5}, 3 d^{6}\) and \(3 d^{8}\) respectively. Which of the following complex will show minimum paramagnetic behaviour?
\(\mathrm{Cr}^{2+}, \mathrm{Mn}^{2+}, \mathrm{Fe}^{2+}\) and \(\mathrm{Ni}^{2+}\) have \(4,5,4\) and 2 unpaired electrons. No of unpaired electrons \(\alpha \) paramagnetism.
CHXII09:COORDINATION COMPOUNDS
322297
Match Column I with Column II and choose the correct combination from the options given.
Larger the number of unpaired electrons, the more it has paramagnetism. The configuration of \(\mathrm{Cr}^{3+}\left(3 d^{3}\right), \mathrm{Fe}^{2+}\left(3 d^{6}\right), \mathrm{Cu}^{2+}\left(3 d^{9}\right)\) and \(\mathrm{Zn}^{2+}\left(3 d^{10}\right)\) are outer orbital complex ions. Hence \(\mathrm{Fe}^{2+}\) has maximum unpaired electrons.
CHXII09:COORDINATION COMPOUNDS
322295
The magnetic moment is measured in Bohr Magneton (B.M.). Spin only magnetic moment of Fe in \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\) and \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}\) complexes respectively is
1 5.92 B.M. and 1.732 B.M.
2 3.87 B.M. and 1.732 B.M.
3 4.89 B.M. and 6.92 B.M.
4 6.92 B.M. in both.
Explanation:
In both the complexes, Fe is in \( + \,3\) oxidation state. \(\mathrm{CN}^{-}\) is a strong field ligand, but \(\mathrm{H}_{2} \mathrm{O}\) is a weak field ligand.
There are 5 unpaired electrons. Spin only magnetic moment value \(=\sqrt{\mathrm{n}(\mathrm{n}+2)}=\sqrt{5(5+2)}=\sqrt{35}=5.92\) B.M.
There is only one unpaired electron. Spin only magnetic moment value \(=\sqrt{\mathrm{n}(\mathrm{n}+2)}=\sqrt{1(1+2)}=\sqrt{3}=1.732\) B.M. Thus, option (1) is correct.
JEE - 2023
CHXII09:COORDINATION COMPOUNDS
322296
The \(d\)-electronic configuration of \(\mathrm{Cr}^{2+}, \mathrm{Mn}^{2+}, \mathrm{Fe}^{2+} \mathrm{Ni}^{2+}\) are \(3 d^{4}, 3 d^{5}, 3 d^{6}\) and \(3 d^{8}\) respectively. Which of the following complex will show minimum paramagnetic behaviour?
\(\mathrm{Cr}^{2+}, \mathrm{Mn}^{2+}, \mathrm{Fe}^{2+}\) and \(\mathrm{Ni}^{2+}\) have \(4,5,4\) and 2 unpaired electrons. No of unpaired electrons \(\alpha \) paramagnetism.
CHXII09:COORDINATION COMPOUNDS
322297
Match Column I with Column II and choose the correct combination from the options given.
Larger the number of unpaired electrons, the more it has paramagnetism. The configuration of \(\mathrm{Cr}^{3+}\left(3 d^{3}\right), \mathrm{Fe}^{2+}\left(3 d^{6}\right), \mathrm{Cu}^{2+}\left(3 d^{9}\right)\) and \(\mathrm{Zn}^{2+}\left(3 d^{10}\right)\) are outer orbital complex ions. Hence \(\mathrm{Fe}^{2+}\) has maximum unpaired electrons.
CHXII09:COORDINATION COMPOUNDS
322295
The magnetic moment is measured in Bohr Magneton (B.M.). Spin only magnetic moment of Fe in \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\) and \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}\) complexes respectively is
1 5.92 B.M. and 1.732 B.M.
2 3.87 B.M. and 1.732 B.M.
3 4.89 B.M. and 6.92 B.M.
4 6.92 B.M. in both.
Explanation:
In both the complexes, Fe is in \( + \,3\) oxidation state. \(\mathrm{CN}^{-}\) is a strong field ligand, but \(\mathrm{H}_{2} \mathrm{O}\) is a weak field ligand.
There are 5 unpaired electrons. Spin only magnetic moment value \(=\sqrt{\mathrm{n}(\mathrm{n}+2)}=\sqrt{5(5+2)}=\sqrt{35}=5.92\) B.M.
There is only one unpaired electron. Spin only magnetic moment value \(=\sqrt{\mathrm{n}(\mathrm{n}+2)}=\sqrt{1(1+2)}=\sqrt{3}=1.732\) B.M. Thus, option (1) is correct.
JEE - 2023
CHXII09:COORDINATION COMPOUNDS
322296
The \(d\)-electronic configuration of \(\mathrm{Cr}^{2+}, \mathrm{Mn}^{2+}, \mathrm{Fe}^{2+} \mathrm{Ni}^{2+}\) are \(3 d^{4}, 3 d^{5}, 3 d^{6}\) and \(3 d^{8}\) respectively. Which of the following complex will show minimum paramagnetic behaviour?
\(\mathrm{Cr}^{2+}, \mathrm{Mn}^{2+}, \mathrm{Fe}^{2+}\) and \(\mathrm{Ni}^{2+}\) have \(4,5,4\) and 2 unpaired electrons. No of unpaired electrons \(\alpha \) paramagnetism.
CHXII09:COORDINATION COMPOUNDS
322297
Match Column I with Column II and choose the correct combination from the options given.
Larger the number of unpaired electrons, the more it has paramagnetism. The configuration of \(\mathrm{Cr}^{3+}\left(3 d^{3}\right), \mathrm{Fe}^{2+}\left(3 d^{6}\right), \mathrm{Cu}^{2+}\left(3 d^{9}\right)\) and \(\mathrm{Zn}^{2+}\left(3 d^{10}\right)\) are outer orbital complex ions. Hence \(\mathrm{Fe}^{2+}\) has maximum unpaired electrons.
CHXII09:COORDINATION COMPOUNDS
322295
The magnetic moment is measured in Bohr Magneton (B.M.). Spin only magnetic moment of Fe in \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\) and \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}\) complexes respectively is
1 5.92 B.M. and 1.732 B.M.
2 3.87 B.M. and 1.732 B.M.
3 4.89 B.M. and 6.92 B.M.
4 6.92 B.M. in both.
Explanation:
In both the complexes, Fe is in \( + \,3\) oxidation state. \(\mathrm{CN}^{-}\) is a strong field ligand, but \(\mathrm{H}_{2} \mathrm{O}\) is a weak field ligand.
There are 5 unpaired electrons. Spin only magnetic moment value \(=\sqrt{\mathrm{n}(\mathrm{n}+2)}=\sqrt{5(5+2)}=\sqrt{35}=5.92\) B.M.
There is only one unpaired electron. Spin only magnetic moment value \(=\sqrt{\mathrm{n}(\mathrm{n}+2)}=\sqrt{1(1+2)}=\sqrt{3}=1.732\) B.M. Thus, option (1) is correct.
JEE - 2023
CHXII09:COORDINATION COMPOUNDS
322296
The \(d\)-electronic configuration of \(\mathrm{Cr}^{2+}, \mathrm{Mn}^{2+}, \mathrm{Fe}^{2+} \mathrm{Ni}^{2+}\) are \(3 d^{4}, 3 d^{5}, 3 d^{6}\) and \(3 d^{8}\) respectively. Which of the following complex will show minimum paramagnetic behaviour?
\(\mathrm{Cr}^{2+}, \mathrm{Mn}^{2+}, \mathrm{Fe}^{2+}\) and \(\mathrm{Ni}^{2+}\) have \(4,5,4\) and 2 unpaired electrons. No of unpaired electrons \(\alpha \) paramagnetism.
CHXII09:COORDINATION COMPOUNDS
322297
Match Column I with Column II and choose the correct combination from the options given.
