\(\mathrm{Fe}^{+3}-3 \mathrm{~d}^{5}\) (5 unpaired electrons). Fe will be attracted in magnetic field so it shows increase in weight.
CHXII08:THE D- & F-BLOCK ELEMENTS
321488
The highest magnetic moment is shown by the transition metal ion with the configuration
1 \(\mathrm{3 d^{2}}\)
2 \(\mathrm{3 d^{5}}\)
3 \(\mathrm{3 d^{7}}\)
4 \(\mathrm{3 d^{9}}\)
Explanation:
Magnetic moment \({\rm{ = }}\sqrt {{\rm{n(n + 2)}}} \,{\rm{BM}}\) where \(\mathrm{\mathrm{n}=}\) number of unpaired electrons. Since \(\mathrm{3 d^{5}}\) has maximum number of unpaired electrons, its magnetic moment will be highest.
KCET - 2006
CHXII08:THE D- & F-BLOCK ELEMENTS
321489
Among the following, the incorrect statement is
1 The fourth period atom having a total of six ' \(\mathrm{\mathrm{d}}\) ' electrons is \(\mathrm{\mathrm{Fe}}\)
2 \(\mathrm{\mathrm{Ca}^{2+}}\) has a smaller radius than \(\mathrm{\mathrm{K}^{+}}\), because it has a higher nuclear charge
3 The magnetic moment for \(\mathrm{\mathrm{Co}^{3+}}\) is \({\rm{4}}{\rm{.90}}\,\,{\rm{BM}}\)
4 Among \(\mathrm{\mathrm{Fe}^{2+}}\) and \(\mathrm{\mathrm{Fe}^{3+}}\), the smaller ion is less paramagnetic.
Explanation:
\(\mathrm{\mathrm{Fe}^{3+}}\) is smaller in size. Its electronic configuration is \(\mathrm{[A r] 3 d^{5}}\). It has 5 unpaired electrons and thus is more paramagnetic than \(\mathrm{\mathrm{Fe}^{2+}}\) ion (which has only 4 unpaired electrons)
CHXII08:THE D- & F-BLOCK ELEMENTS
321490
For which of the following pairs, magnetic moment is same?
1 \(\mathrm{MnCl}_{2}, \mathrm{CuSO}_{4}\)
2 \(\mathrm{CuCl}_{2}, \mathrm{TiCl}_{3}\)
3 \(\mathrm{TiO}_{2}, \mathrm{CuSO}_{4}\)
4 \(\mathrm{TiCl}_{3}, \mathrm{NiCl}_{2}\)
Explanation:
Species having the same number of unpaired electrons, have same magnetic moment. (1) \(\mathrm{MnCl}_{2} \Rightarrow \mathrm{Mn}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^{5}, 4 \mathrm{~s}^{0}\) (five unpaired electrons) \(\mathrm{CuSO}_{4} \Rightarrow \mathrm{Cu}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^{9}, 4 \mathrm{~s}^{0}\) (one unpaired electron) (2) \(\mathrm{CuCl}_{2} \Rightarrow \mathrm{Cu}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^{9}, 4 \mathrm{~s}^{0}\) (one unpaired electron) \(\mathrm{TiCl}_{3} \Rightarrow \mathrm{Ti}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^{1}, 4 \mathrm{~s}^{0}\) (one unpaired electron) (3) \(\mathrm{TiO}_{2} \Rightarrow \mathrm{Ti}^{4+}=[\mathrm{Ar}] 3 \mathrm{~d}^{0}, 4 \mathrm{~s}^{0}\) (no unpaired electron) \(\mathrm{CuSO}_{4} \Rightarrow \mathrm{Cu}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^{9}, 4 \mathrm{~s}^{0}\) (one unpaired electron) (4) \(\mathrm{TiCl}_{3} \Rightarrow \mathrm{Ti}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^{1}, 4 \mathrm{~s}^{0}\) (one unpaired electron) \(\mathrm{NiCl}_{2} \Rightarrow \mathrm{Ni}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^{8}, 4 \mathrm{~s}^{0}\) (two unpaired electrons) \(\therefore \mathrm{CuCl}_{2}\) and \(\mathrm{TiCl}_{3}\) will have same magnetic moment. \(\therefore \mu = \sqrt {1(1 + 2)} = \sqrt {\text{3}} = 1.73\,\,{\text{BM}}\)
\(\mathrm{Fe}^{+3}-3 \mathrm{~d}^{5}\) (5 unpaired electrons). Fe will be attracted in magnetic field so it shows increase in weight.
CHXII08:THE D- & F-BLOCK ELEMENTS
321488
The highest magnetic moment is shown by the transition metal ion with the configuration
1 \(\mathrm{3 d^{2}}\)
2 \(\mathrm{3 d^{5}}\)
3 \(\mathrm{3 d^{7}}\)
4 \(\mathrm{3 d^{9}}\)
Explanation:
Magnetic moment \({\rm{ = }}\sqrt {{\rm{n(n + 2)}}} \,{\rm{BM}}\) where \(\mathrm{\mathrm{n}=}\) number of unpaired electrons. Since \(\mathrm{3 d^{5}}\) has maximum number of unpaired electrons, its magnetic moment will be highest.
KCET - 2006
CHXII08:THE D- & F-BLOCK ELEMENTS
321489
Among the following, the incorrect statement is
1 The fourth period atom having a total of six ' \(\mathrm{\mathrm{d}}\) ' electrons is \(\mathrm{\mathrm{Fe}}\)
2 \(\mathrm{\mathrm{Ca}^{2+}}\) has a smaller radius than \(\mathrm{\mathrm{K}^{+}}\), because it has a higher nuclear charge
3 The magnetic moment for \(\mathrm{\mathrm{Co}^{3+}}\) is \({\rm{4}}{\rm{.90}}\,\,{\rm{BM}}\)
4 Among \(\mathrm{\mathrm{Fe}^{2+}}\) and \(\mathrm{\mathrm{Fe}^{3+}}\), the smaller ion is less paramagnetic.
Explanation:
\(\mathrm{\mathrm{Fe}^{3+}}\) is smaller in size. Its electronic configuration is \(\mathrm{[A r] 3 d^{5}}\). It has 5 unpaired electrons and thus is more paramagnetic than \(\mathrm{\mathrm{Fe}^{2+}}\) ion (which has only 4 unpaired electrons)
CHXII08:THE D- & F-BLOCK ELEMENTS
321490
For which of the following pairs, magnetic moment is same?
1 \(\mathrm{MnCl}_{2}, \mathrm{CuSO}_{4}\)
2 \(\mathrm{CuCl}_{2}, \mathrm{TiCl}_{3}\)
3 \(\mathrm{TiO}_{2}, \mathrm{CuSO}_{4}\)
4 \(\mathrm{TiCl}_{3}, \mathrm{NiCl}_{2}\)
Explanation:
Species having the same number of unpaired electrons, have same magnetic moment. (1) \(\mathrm{MnCl}_{2} \Rightarrow \mathrm{Mn}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^{5}, 4 \mathrm{~s}^{0}\) (five unpaired electrons) \(\mathrm{CuSO}_{4} \Rightarrow \mathrm{Cu}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^{9}, 4 \mathrm{~s}^{0}\) (one unpaired electron) (2) \(\mathrm{CuCl}_{2} \Rightarrow \mathrm{Cu}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^{9}, 4 \mathrm{~s}^{0}\) (one unpaired electron) \(\mathrm{TiCl}_{3} \Rightarrow \mathrm{Ti}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^{1}, 4 \mathrm{~s}^{0}\) (one unpaired electron) (3) \(\mathrm{TiO}_{2} \Rightarrow \mathrm{Ti}^{4+}=[\mathrm{Ar}] 3 \mathrm{~d}^{0}, 4 \mathrm{~s}^{0}\) (no unpaired electron) \(\mathrm{CuSO}_{4} \Rightarrow \mathrm{Cu}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^{9}, 4 \mathrm{~s}^{0}\) (one unpaired electron) (4) \(\mathrm{TiCl}_{3} \Rightarrow \mathrm{Ti}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^{1}, 4 \mathrm{~s}^{0}\) (one unpaired electron) \(\mathrm{NiCl}_{2} \Rightarrow \mathrm{Ni}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^{8}, 4 \mathrm{~s}^{0}\) (two unpaired electrons) \(\therefore \mathrm{CuCl}_{2}\) and \(\mathrm{TiCl}_{3}\) will have same magnetic moment. \(\therefore \mu = \sqrt {1(1 + 2)} = \sqrt {\text{3}} = 1.73\,\,{\text{BM}}\)
\(\mathrm{Fe}^{+3}-3 \mathrm{~d}^{5}\) (5 unpaired electrons). Fe will be attracted in magnetic field so it shows increase in weight.
CHXII08:THE D- & F-BLOCK ELEMENTS
321488
The highest magnetic moment is shown by the transition metal ion with the configuration
1 \(\mathrm{3 d^{2}}\)
2 \(\mathrm{3 d^{5}}\)
3 \(\mathrm{3 d^{7}}\)
4 \(\mathrm{3 d^{9}}\)
Explanation:
Magnetic moment \({\rm{ = }}\sqrt {{\rm{n(n + 2)}}} \,{\rm{BM}}\) where \(\mathrm{\mathrm{n}=}\) number of unpaired electrons. Since \(\mathrm{3 d^{5}}\) has maximum number of unpaired electrons, its magnetic moment will be highest.
KCET - 2006
CHXII08:THE D- & F-BLOCK ELEMENTS
321489
Among the following, the incorrect statement is
1 The fourth period atom having a total of six ' \(\mathrm{\mathrm{d}}\) ' electrons is \(\mathrm{\mathrm{Fe}}\)
2 \(\mathrm{\mathrm{Ca}^{2+}}\) has a smaller radius than \(\mathrm{\mathrm{K}^{+}}\), because it has a higher nuclear charge
3 The magnetic moment for \(\mathrm{\mathrm{Co}^{3+}}\) is \({\rm{4}}{\rm{.90}}\,\,{\rm{BM}}\)
4 Among \(\mathrm{\mathrm{Fe}^{2+}}\) and \(\mathrm{\mathrm{Fe}^{3+}}\), the smaller ion is less paramagnetic.
Explanation:
\(\mathrm{\mathrm{Fe}^{3+}}\) is smaller in size. Its electronic configuration is \(\mathrm{[A r] 3 d^{5}}\). It has 5 unpaired electrons and thus is more paramagnetic than \(\mathrm{\mathrm{Fe}^{2+}}\) ion (which has only 4 unpaired electrons)
CHXII08:THE D- & F-BLOCK ELEMENTS
321490
For which of the following pairs, magnetic moment is same?
1 \(\mathrm{MnCl}_{2}, \mathrm{CuSO}_{4}\)
2 \(\mathrm{CuCl}_{2}, \mathrm{TiCl}_{3}\)
3 \(\mathrm{TiO}_{2}, \mathrm{CuSO}_{4}\)
4 \(\mathrm{TiCl}_{3}, \mathrm{NiCl}_{2}\)
Explanation:
Species having the same number of unpaired electrons, have same magnetic moment. (1) \(\mathrm{MnCl}_{2} \Rightarrow \mathrm{Mn}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^{5}, 4 \mathrm{~s}^{0}\) (five unpaired electrons) \(\mathrm{CuSO}_{4} \Rightarrow \mathrm{Cu}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^{9}, 4 \mathrm{~s}^{0}\) (one unpaired electron) (2) \(\mathrm{CuCl}_{2} \Rightarrow \mathrm{Cu}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^{9}, 4 \mathrm{~s}^{0}\) (one unpaired electron) \(\mathrm{TiCl}_{3} \Rightarrow \mathrm{Ti}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^{1}, 4 \mathrm{~s}^{0}\) (one unpaired electron) (3) \(\mathrm{TiO}_{2} \Rightarrow \mathrm{Ti}^{4+}=[\mathrm{Ar}] 3 \mathrm{~d}^{0}, 4 \mathrm{~s}^{0}\) (no unpaired electron) \(\mathrm{CuSO}_{4} \Rightarrow \mathrm{Cu}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^{9}, 4 \mathrm{~s}^{0}\) (one unpaired electron) (4) \(\mathrm{TiCl}_{3} \Rightarrow \mathrm{Ti}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^{1}, 4 \mathrm{~s}^{0}\) (one unpaired electron) \(\mathrm{NiCl}_{2} \Rightarrow \mathrm{Ni}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^{8}, 4 \mathrm{~s}^{0}\) (two unpaired electrons) \(\therefore \mathrm{CuCl}_{2}\) and \(\mathrm{TiCl}_{3}\) will have same magnetic moment. \(\therefore \mu = \sqrt {1(1 + 2)} = \sqrt {\text{3}} = 1.73\,\,{\text{BM}}\)
\(\mathrm{Fe}^{+3}-3 \mathrm{~d}^{5}\) (5 unpaired electrons). Fe will be attracted in magnetic field so it shows increase in weight.
CHXII08:THE D- & F-BLOCK ELEMENTS
321488
The highest magnetic moment is shown by the transition metal ion with the configuration
1 \(\mathrm{3 d^{2}}\)
2 \(\mathrm{3 d^{5}}\)
3 \(\mathrm{3 d^{7}}\)
4 \(\mathrm{3 d^{9}}\)
Explanation:
Magnetic moment \({\rm{ = }}\sqrt {{\rm{n(n + 2)}}} \,{\rm{BM}}\) where \(\mathrm{\mathrm{n}=}\) number of unpaired electrons. Since \(\mathrm{3 d^{5}}\) has maximum number of unpaired electrons, its magnetic moment will be highest.
KCET - 2006
CHXII08:THE D- & F-BLOCK ELEMENTS
321489
Among the following, the incorrect statement is
1 The fourth period atom having a total of six ' \(\mathrm{\mathrm{d}}\) ' electrons is \(\mathrm{\mathrm{Fe}}\)
2 \(\mathrm{\mathrm{Ca}^{2+}}\) has a smaller radius than \(\mathrm{\mathrm{K}^{+}}\), because it has a higher nuclear charge
3 The magnetic moment for \(\mathrm{\mathrm{Co}^{3+}}\) is \({\rm{4}}{\rm{.90}}\,\,{\rm{BM}}\)
4 Among \(\mathrm{\mathrm{Fe}^{2+}}\) and \(\mathrm{\mathrm{Fe}^{3+}}\), the smaller ion is less paramagnetic.
Explanation:
\(\mathrm{\mathrm{Fe}^{3+}}\) is smaller in size. Its electronic configuration is \(\mathrm{[A r] 3 d^{5}}\). It has 5 unpaired electrons and thus is more paramagnetic than \(\mathrm{\mathrm{Fe}^{2+}}\) ion (which has only 4 unpaired electrons)
CHXII08:THE D- & F-BLOCK ELEMENTS
321490
For which of the following pairs, magnetic moment is same?
1 \(\mathrm{MnCl}_{2}, \mathrm{CuSO}_{4}\)
2 \(\mathrm{CuCl}_{2}, \mathrm{TiCl}_{3}\)
3 \(\mathrm{TiO}_{2}, \mathrm{CuSO}_{4}\)
4 \(\mathrm{TiCl}_{3}, \mathrm{NiCl}_{2}\)
Explanation:
Species having the same number of unpaired electrons, have same magnetic moment. (1) \(\mathrm{MnCl}_{2} \Rightarrow \mathrm{Mn}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^{5}, 4 \mathrm{~s}^{0}\) (five unpaired electrons) \(\mathrm{CuSO}_{4} \Rightarrow \mathrm{Cu}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^{9}, 4 \mathrm{~s}^{0}\) (one unpaired electron) (2) \(\mathrm{CuCl}_{2} \Rightarrow \mathrm{Cu}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^{9}, 4 \mathrm{~s}^{0}\) (one unpaired electron) \(\mathrm{TiCl}_{3} \Rightarrow \mathrm{Ti}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^{1}, 4 \mathrm{~s}^{0}\) (one unpaired electron) (3) \(\mathrm{TiO}_{2} \Rightarrow \mathrm{Ti}^{4+}=[\mathrm{Ar}] 3 \mathrm{~d}^{0}, 4 \mathrm{~s}^{0}\) (no unpaired electron) \(\mathrm{CuSO}_{4} \Rightarrow \mathrm{Cu}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^{9}, 4 \mathrm{~s}^{0}\) (one unpaired electron) (4) \(\mathrm{TiCl}_{3} \Rightarrow \mathrm{Ti}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^{1}, 4 \mathrm{~s}^{0}\) (one unpaired electron) \(\mathrm{NiCl}_{2} \Rightarrow \mathrm{Ni}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^{8}, 4 \mathrm{~s}^{0}\) (two unpaired electrons) \(\therefore \mathrm{CuCl}_{2}\) and \(\mathrm{TiCl}_{3}\) will have same magnetic moment. \(\therefore \mu = \sqrt {1(1 + 2)} = \sqrt {\text{3}} = 1.73\,\,{\text{BM}}\)
