321496
Chloro compound of Vanadium has only spin magnetic moment of \({\rm{1}}{\rm{.73}}{\mkern 1mu} {\mkern 1mu} {\rm{BM}}\). This Vanadium chloride has the formula:
1 \(\mathrm{\mathrm{VCl}_{4}}\)
2 \(\mathrm{\mathrm{VCl}_{3}}\)
3 \(\mathrm{\mathrm{VCl}_{12}}\)
4 \(\mathrm{\mathrm{VCl}_{5}}\)
Explanation:
If the magnetic moment is \({\rm{1}}{\rm{.73}}\,{\rm{BM}}\) then the state of Vanadium is \(\mathrm{\mathrm{V}^{4+}}\) with one unpaired electron. Thus, the formula of the chloride is \({\rm{VC}}{{\rm{l}}_{\rm{4}}}\).
JEE - 2014
CHXII08:THE D- & F-BLOCK ELEMENTS
321497
Which of the following weighs less when weighed in magnetic field ?
1 \(\mathrm{\mathrm{VCl}_{3}}\)
2 \(\mathrm{\mathrm{ScCl}_{3}}\)
3 \(\mathrm{\mathrm{TiCl}_{3}}\)
4 \(\mathrm{\mathrm{FeCl}_{3}}\)
Explanation:
\(\mathrm{\mathrm{ScCl}_{3} \rightarrow \mathrm{Sc}^{3+}+3 \mathrm{Cl}^{-}, \mathrm{Sc}^{3+}=[\mathrm{Ar}] 3 d^{0}}\) Since there is no unpaired electron, it will show diamagnetic character and will be repelled. Hence, it will weigh less.
CHXII08:THE D- & F-BLOCK ELEMENTS
321498
In a transition series, with the increase in atomic number, the paramagnetism
1 increases gradually
2 decreases gradually
3 first increases to a maximum and then decreases
4 first decreases to a minimum and then increases.
Explanation:
In transition series, paramagnetism first increases due to increase in the number of unpaired electrons. Beyond manganese, number of unpaired electrons decreases. So, paramagnetism decreases.
KCET - 2013
CHXII08:THE D- & F-BLOCK ELEMENTS
321499
The spin only magnetic moment of \(\mathrm{\mathrm{Mn}^{4+}}\) ion is approximately
1 3 B.M.
2 6 B.M.
3 4 B.M.
4 5 B.M.
Explanation:
\(\mathrm{M n^{4+} \rightarrow 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{3}}\) Magnetic moment \(\mathrm{=\sqrt{n(n+2)}}\) B.M. Number of unpaired electrons, \(\mathrm{\mathrm{n}=3}\) Magnetic moment \(\mathrm{=}\) \(\rm{\sqrt{3(3+2)}=\sqrt{3(5)}=\sqrt{15}=3.89=4 \text { B.M }}\)
321496
Chloro compound of Vanadium has only spin magnetic moment of \({\rm{1}}{\rm{.73}}{\mkern 1mu} {\mkern 1mu} {\rm{BM}}\). This Vanadium chloride has the formula:
1 \(\mathrm{\mathrm{VCl}_{4}}\)
2 \(\mathrm{\mathrm{VCl}_{3}}\)
3 \(\mathrm{\mathrm{VCl}_{12}}\)
4 \(\mathrm{\mathrm{VCl}_{5}}\)
Explanation:
If the magnetic moment is \({\rm{1}}{\rm{.73}}\,{\rm{BM}}\) then the state of Vanadium is \(\mathrm{\mathrm{V}^{4+}}\) with one unpaired electron. Thus, the formula of the chloride is \({\rm{VC}}{{\rm{l}}_{\rm{4}}}\).
JEE - 2014
CHXII08:THE D- & F-BLOCK ELEMENTS
321497
Which of the following weighs less when weighed in magnetic field ?
1 \(\mathrm{\mathrm{VCl}_{3}}\)
2 \(\mathrm{\mathrm{ScCl}_{3}}\)
3 \(\mathrm{\mathrm{TiCl}_{3}}\)
4 \(\mathrm{\mathrm{FeCl}_{3}}\)
Explanation:
\(\mathrm{\mathrm{ScCl}_{3} \rightarrow \mathrm{Sc}^{3+}+3 \mathrm{Cl}^{-}, \mathrm{Sc}^{3+}=[\mathrm{Ar}] 3 d^{0}}\) Since there is no unpaired electron, it will show diamagnetic character and will be repelled. Hence, it will weigh less.
CHXII08:THE D- & F-BLOCK ELEMENTS
321498
In a transition series, with the increase in atomic number, the paramagnetism
1 increases gradually
2 decreases gradually
3 first increases to a maximum and then decreases
4 first decreases to a minimum and then increases.
Explanation:
In transition series, paramagnetism first increases due to increase in the number of unpaired electrons. Beyond manganese, number of unpaired electrons decreases. So, paramagnetism decreases.
KCET - 2013
CHXII08:THE D- & F-BLOCK ELEMENTS
321499
The spin only magnetic moment of \(\mathrm{\mathrm{Mn}^{4+}}\) ion is approximately
1 3 B.M.
2 6 B.M.
3 4 B.M.
4 5 B.M.
Explanation:
\(\mathrm{M n^{4+} \rightarrow 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{3}}\) Magnetic moment \(\mathrm{=\sqrt{n(n+2)}}\) B.M. Number of unpaired electrons, \(\mathrm{\mathrm{n}=3}\) Magnetic moment \(\mathrm{=}\) \(\rm{\sqrt{3(3+2)}=\sqrt{3(5)}=\sqrt{15}=3.89=4 \text { B.M }}\)
321496
Chloro compound of Vanadium has only spin magnetic moment of \({\rm{1}}{\rm{.73}}{\mkern 1mu} {\mkern 1mu} {\rm{BM}}\). This Vanadium chloride has the formula:
1 \(\mathrm{\mathrm{VCl}_{4}}\)
2 \(\mathrm{\mathrm{VCl}_{3}}\)
3 \(\mathrm{\mathrm{VCl}_{12}}\)
4 \(\mathrm{\mathrm{VCl}_{5}}\)
Explanation:
If the magnetic moment is \({\rm{1}}{\rm{.73}}\,{\rm{BM}}\) then the state of Vanadium is \(\mathrm{\mathrm{V}^{4+}}\) with one unpaired electron. Thus, the formula of the chloride is \({\rm{VC}}{{\rm{l}}_{\rm{4}}}\).
JEE - 2014
CHXII08:THE D- & F-BLOCK ELEMENTS
321497
Which of the following weighs less when weighed in magnetic field ?
1 \(\mathrm{\mathrm{VCl}_{3}}\)
2 \(\mathrm{\mathrm{ScCl}_{3}}\)
3 \(\mathrm{\mathrm{TiCl}_{3}}\)
4 \(\mathrm{\mathrm{FeCl}_{3}}\)
Explanation:
\(\mathrm{\mathrm{ScCl}_{3} \rightarrow \mathrm{Sc}^{3+}+3 \mathrm{Cl}^{-}, \mathrm{Sc}^{3+}=[\mathrm{Ar}] 3 d^{0}}\) Since there is no unpaired electron, it will show diamagnetic character and will be repelled. Hence, it will weigh less.
CHXII08:THE D- & F-BLOCK ELEMENTS
321498
In a transition series, with the increase in atomic number, the paramagnetism
1 increases gradually
2 decreases gradually
3 first increases to a maximum and then decreases
4 first decreases to a minimum and then increases.
Explanation:
In transition series, paramagnetism first increases due to increase in the number of unpaired electrons. Beyond manganese, number of unpaired electrons decreases. So, paramagnetism decreases.
KCET - 2013
CHXII08:THE D- & F-BLOCK ELEMENTS
321499
The spin only magnetic moment of \(\mathrm{\mathrm{Mn}^{4+}}\) ion is approximately
1 3 B.M.
2 6 B.M.
3 4 B.M.
4 5 B.M.
Explanation:
\(\mathrm{M n^{4+} \rightarrow 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{3}}\) Magnetic moment \(\mathrm{=\sqrt{n(n+2)}}\) B.M. Number of unpaired electrons, \(\mathrm{\mathrm{n}=3}\) Magnetic moment \(\mathrm{=}\) \(\rm{\sqrt{3(3+2)}=\sqrt{3(5)}=\sqrt{15}=3.89=4 \text { B.M }}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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CHXII08:THE D- & F-BLOCK ELEMENTS
321496
Chloro compound of Vanadium has only spin magnetic moment of \({\rm{1}}{\rm{.73}}{\mkern 1mu} {\mkern 1mu} {\rm{BM}}\). This Vanadium chloride has the formula:
1 \(\mathrm{\mathrm{VCl}_{4}}\)
2 \(\mathrm{\mathrm{VCl}_{3}}\)
3 \(\mathrm{\mathrm{VCl}_{12}}\)
4 \(\mathrm{\mathrm{VCl}_{5}}\)
Explanation:
If the magnetic moment is \({\rm{1}}{\rm{.73}}\,{\rm{BM}}\) then the state of Vanadium is \(\mathrm{\mathrm{V}^{4+}}\) with one unpaired electron. Thus, the formula of the chloride is \({\rm{VC}}{{\rm{l}}_{\rm{4}}}\).
JEE - 2014
CHXII08:THE D- & F-BLOCK ELEMENTS
321497
Which of the following weighs less when weighed in magnetic field ?
1 \(\mathrm{\mathrm{VCl}_{3}}\)
2 \(\mathrm{\mathrm{ScCl}_{3}}\)
3 \(\mathrm{\mathrm{TiCl}_{3}}\)
4 \(\mathrm{\mathrm{FeCl}_{3}}\)
Explanation:
\(\mathrm{\mathrm{ScCl}_{3} \rightarrow \mathrm{Sc}^{3+}+3 \mathrm{Cl}^{-}, \mathrm{Sc}^{3+}=[\mathrm{Ar}] 3 d^{0}}\) Since there is no unpaired electron, it will show diamagnetic character and will be repelled. Hence, it will weigh less.
CHXII08:THE D- & F-BLOCK ELEMENTS
321498
In a transition series, with the increase in atomic number, the paramagnetism
1 increases gradually
2 decreases gradually
3 first increases to a maximum and then decreases
4 first decreases to a minimum and then increases.
Explanation:
In transition series, paramagnetism first increases due to increase in the number of unpaired electrons. Beyond manganese, number of unpaired electrons decreases. So, paramagnetism decreases.
KCET - 2013
CHXII08:THE D- & F-BLOCK ELEMENTS
321499
The spin only magnetic moment of \(\mathrm{\mathrm{Mn}^{4+}}\) ion is approximately
1 3 B.M.
2 6 B.M.
3 4 B.M.
4 5 B.M.
Explanation:
\(\mathrm{M n^{4+} \rightarrow 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{3}}\) Magnetic moment \(\mathrm{=\sqrt{n(n+2)}}\) B.M. Number of unpaired electrons, \(\mathrm{\mathrm{n}=3}\) Magnetic moment \(\mathrm{=}\) \(\rm{\sqrt{3(3+2)}=\sqrt{3(5)}=\sqrt{15}=3.89=4 \text { B.M }}\)
