More is the number of unpaired electrons, greater is the paramagnetism and thus, higher is the attraction towards magnetic field. The number of unpaired electrons in \(\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+},\left[\mathrm{Co}(\mathrm{en})_{3}\right]^{3+},\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) and \(\left[\mathrm{Zn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) are respectively \(3,0,2,0\). So, \(\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) will exhibit maximum attraction towards magnetic field.
JEE - 2023
CHXII09:COORDINATION COMPOUNDS
322313
\(\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}\) is paramagnetic while \(\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}\) is diamagnetic because
1 \(\mathrm{Cr}^{3+}\) has \(\mathrm{d}^{3}\) configuration while \(\mathrm{Ni}^{2+}\) had \(\mathrm{d}^{8}\) configuration
2 \(\mathrm{Cr}^{3+}\) form \(\mathrm{d}^{2} \mathrm{sp}^{3}\) hybridisation while \(\mathrm{Ni}^{2+}\) forms dsp \({ }^{2}\) hybridisation
3 There are 3, 0 unpaired electrons in \(\mathrm{Cr}^{3+}\) and \(\mathrm{Ni}^{2+}\)
4 All the above
Explanation:
Conceptual Questions
CHXII09:COORDINATION COMPOUNDS
322314
Which of the following complexes exhibits the highest paramagnetic behaviour?
\(\mathrm{Fe}^{2+}, \mathrm{Co}^{5+}, \mathrm{Ti}^{3+}\) and \(\mathrm{V}^{3+}\) have \(4,4,1,2\) unpaired electrons respectively. The pairing leads \(\mathrm{Fe}^{2+}\) with no unpaired electron.
More is the number of unpaired electrons, greater is the paramagnetism and thus, higher is the attraction towards magnetic field. The number of unpaired electrons in \(\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+},\left[\mathrm{Co}(\mathrm{en})_{3}\right]^{3+},\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) and \(\left[\mathrm{Zn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) are respectively \(3,0,2,0\). So, \(\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) will exhibit maximum attraction towards magnetic field.
JEE - 2023
CHXII09:COORDINATION COMPOUNDS
322313
\(\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}\) is paramagnetic while \(\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}\) is diamagnetic because
1 \(\mathrm{Cr}^{3+}\) has \(\mathrm{d}^{3}\) configuration while \(\mathrm{Ni}^{2+}\) had \(\mathrm{d}^{8}\) configuration
2 \(\mathrm{Cr}^{3+}\) form \(\mathrm{d}^{2} \mathrm{sp}^{3}\) hybridisation while \(\mathrm{Ni}^{2+}\) forms dsp \({ }^{2}\) hybridisation
3 There are 3, 0 unpaired electrons in \(\mathrm{Cr}^{3+}\) and \(\mathrm{Ni}^{2+}\)
4 All the above
Explanation:
Conceptual Questions
CHXII09:COORDINATION COMPOUNDS
322314
Which of the following complexes exhibits the highest paramagnetic behaviour?
\(\mathrm{Fe}^{2+}, \mathrm{Co}^{5+}, \mathrm{Ti}^{3+}\) and \(\mathrm{V}^{3+}\) have \(4,4,1,2\) unpaired electrons respectively. The pairing leads \(\mathrm{Fe}^{2+}\) with no unpaired electron.
More is the number of unpaired electrons, greater is the paramagnetism and thus, higher is the attraction towards magnetic field. The number of unpaired electrons in \(\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+},\left[\mathrm{Co}(\mathrm{en})_{3}\right]^{3+},\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) and \(\left[\mathrm{Zn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) are respectively \(3,0,2,0\). So, \(\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) will exhibit maximum attraction towards magnetic field.
JEE - 2023
CHXII09:COORDINATION COMPOUNDS
322313
\(\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}\) is paramagnetic while \(\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}\) is diamagnetic because
1 \(\mathrm{Cr}^{3+}\) has \(\mathrm{d}^{3}\) configuration while \(\mathrm{Ni}^{2+}\) had \(\mathrm{d}^{8}\) configuration
2 \(\mathrm{Cr}^{3+}\) form \(\mathrm{d}^{2} \mathrm{sp}^{3}\) hybridisation while \(\mathrm{Ni}^{2+}\) forms dsp \({ }^{2}\) hybridisation
3 There are 3, 0 unpaired electrons in \(\mathrm{Cr}^{3+}\) and \(\mathrm{Ni}^{2+}\)
4 All the above
Explanation:
Conceptual Questions
CHXII09:COORDINATION COMPOUNDS
322314
Which of the following complexes exhibits the highest paramagnetic behaviour?
\(\mathrm{Fe}^{2+}, \mathrm{Co}^{5+}, \mathrm{Ti}^{3+}\) and \(\mathrm{V}^{3+}\) have \(4,4,1,2\) unpaired electrons respectively. The pairing leads \(\mathrm{Fe}^{2+}\) with no unpaired electron.
More is the number of unpaired electrons, greater is the paramagnetism and thus, higher is the attraction towards magnetic field. The number of unpaired electrons in \(\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+},\left[\mathrm{Co}(\mathrm{en})_{3}\right]^{3+},\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) and \(\left[\mathrm{Zn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) are respectively \(3,0,2,0\). So, \(\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) will exhibit maximum attraction towards magnetic field.
JEE - 2023
CHXII09:COORDINATION COMPOUNDS
322313
\(\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}\) is paramagnetic while \(\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}\) is diamagnetic because
1 \(\mathrm{Cr}^{3+}\) has \(\mathrm{d}^{3}\) configuration while \(\mathrm{Ni}^{2+}\) had \(\mathrm{d}^{8}\) configuration
2 \(\mathrm{Cr}^{3+}\) form \(\mathrm{d}^{2} \mathrm{sp}^{3}\) hybridisation while \(\mathrm{Ni}^{2+}\) forms dsp \({ }^{2}\) hybridisation
3 There are 3, 0 unpaired electrons in \(\mathrm{Cr}^{3+}\) and \(\mathrm{Ni}^{2+}\)
4 All the above
Explanation:
Conceptual Questions
CHXII09:COORDINATION COMPOUNDS
322314
Which of the following complexes exhibits the highest paramagnetic behaviour?
\(\mathrm{Fe}^{2+}, \mathrm{Co}^{5+}, \mathrm{Ti}^{3+}\) and \(\mathrm{V}^{3+}\) have \(4,4,1,2\) unpaired electrons respectively. The pairing leads \(\mathrm{Fe}^{2+}\) with no unpaired electron.
