274067
The spin only magnetic moment $\left(\mu_{s}\right)$ of a complex $\left[\mathrm{MnBr}_{4}\right]^{2-}$ is $5.9 \mathrm{BM}$. The geometry of the complex will be
1 tetrahedral
2 square planar
3 square pyramidal
4 tetragonal
Explanation:
(A) : In $\left[\mathrm{MnBr}_{4}\right]^{2-}, \mathrm{Mn}$ is present in +2 oxidation state. Number of unpaired electron $=5$ Spin only magnetic moment $=\sqrt{5(5+2)}=$ 5.9 B.M. $\mathrm{Br}^{-}$act as weak field ligand so, pairing of d-electrons not possible. Hence, $\left[\mathrm{MnBr}_{4}\right]^{2-}$ forms outer-orbital complex with $\mathrm{sp}^{3}$ hybridisation and geometry is tetrahedral
AMU-2016
COORDINATION COMPOUNDS
274068
The highest magnetic moment is shown by the transition metal ion with outer electronic configuration
1 $d^{9}$
2 $d^{7}$
3 $\mathrm{d}^{5}$
4 $d^{3}$
Explanation:
(C) : More be the number of unpaired electrons, more be the value of magnetic moment. $\mathrm{d}^{5}$ - configuration contains maximum number of unpaired electrons i.e. (5). Thus, show maximum dipole moment as per the following relation. Magnetic moment $\left(\mu_{\text {eff }}\right)=\sqrt{n(n+2)} B M$ where, $\mathrm{n}=$ no. of unpaired electrons.
JCECE - 2016
COORDINATION COMPOUNDS
274091
AB is an ionic solid. If the ratio of ionic radii of $A^{+}$and $B^{-}$is 0.52 , what is the coordination number of $B^{-}$?
1 2
2 3
3 6
4 8
Explanation:
(C) : For the ionic solids, we can write the following radius ratio's of the ions which is given below- | Radius ratio $\mathbf{r}^{+} / \mathbf{r}^{-}$ | Co-ordination number | | :--- | :--- | | $0.155-0.225$ | 3 | | $0.225-0.414$ | 4 | | $0.414-0.732$ | 6 | | $0.732-1.00$ | 8 | Hence, if the ratio of ionic radii is 0.52 , then the coordination number of $\mathrm{B}^{-}$is 6 .
2008]**
COORDINATION COMPOUNDS
274070
Which of the following metal ions has a calculated magnetic moment value of $\sqrt{24}$ BM?
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COORDINATION COMPOUNDS
274067
The spin only magnetic moment $\left(\mu_{s}\right)$ of a complex $\left[\mathrm{MnBr}_{4}\right]^{2-}$ is $5.9 \mathrm{BM}$. The geometry of the complex will be
1 tetrahedral
2 square planar
3 square pyramidal
4 tetragonal
Explanation:
(A) : In $\left[\mathrm{MnBr}_{4}\right]^{2-}, \mathrm{Mn}$ is present in +2 oxidation state. Number of unpaired electron $=5$ Spin only magnetic moment $=\sqrt{5(5+2)}=$ 5.9 B.M. $\mathrm{Br}^{-}$act as weak field ligand so, pairing of d-electrons not possible. Hence, $\left[\mathrm{MnBr}_{4}\right]^{2-}$ forms outer-orbital complex with $\mathrm{sp}^{3}$ hybridisation and geometry is tetrahedral
AMU-2016
COORDINATION COMPOUNDS
274068
The highest magnetic moment is shown by the transition metal ion with outer electronic configuration
1 $d^{9}$
2 $d^{7}$
3 $\mathrm{d}^{5}$
4 $d^{3}$
Explanation:
(C) : More be the number of unpaired electrons, more be the value of magnetic moment. $\mathrm{d}^{5}$ - configuration contains maximum number of unpaired electrons i.e. (5). Thus, show maximum dipole moment as per the following relation. Magnetic moment $\left(\mu_{\text {eff }}\right)=\sqrt{n(n+2)} B M$ where, $\mathrm{n}=$ no. of unpaired electrons.
JCECE - 2016
COORDINATION COMPOUNDS
274091
AB is an ionic solid. If the ratio of ionic radii of $A^{+}$and $B^{-}$is 0.52 , what is the coordination number of $B^{-}$?
1 2
2 3
3 6
4 8
Explanation:
(C) : For the ionic solids, we can write the following radius ratio's of the ions which is given below- | Radius ratio $\mathbf{r}^{+} / \mathbf{r}^{-}$ | Co-ordination number | | :--- | :--- | | $0.155-0.225$ | 3 | | $0.225-0.414$ | 4 | | $0.414-0.732$ | 6 | | $0.732-1.00$ | 8 | Hence, if the ratio of ionic radii is 0.52 , then the coordination number of $\mathrm{B}^{-}$is 6 .
2008]**
COORDINATION COMPOUNDS
274070
Which of the following metal ions has a calculated magnetic moment value of $\sqrt{24}$ BM?
274067
The spin only magnetic moment $\left(\mu_{s}\right)$ of a complex $\left[\mathrm{MnBr}_{4}\right]^{2-}$ is $5.9 \mathrm{BM}$. The geometry of the complex will be
1 tetrahedral
2 square planar
3 square pyramidal
4 tetragonal
Explanation:
(A) : In $\left[\mathrm{MnBr}_{4}\right]^{2-}, \mathrm{Mn}$ is present in +2 oxidation state. Number of unpaired electron $=5$ Spin only magnetic moment $=\sqrt{5(5+2)}=$ 5.9 B.M. $\mathrm{Br}^{-}$act as weak field ligand so, pairing of d-electrons not possible. Hence, $\left[\mathrm{MnBr}_{4}\right]^{2-}$ forms outer-orbital complex with $\mathrm{sp}^{3}$ hybridisation and geometry is tetrahedral
AMU-2016
COORDINATION COMPOUNDS
274068
The highest magnetic moment is shown by the transition metal ion with outer electronic configuration
1 $d^{9}$
2 $d^{7}$
3 $\mathrm{d}^{5}$
4 $d^{3}$
Explanation:
(C) : More be the number of unpaired electrons, more be the value of magnetic moment. $\mathrm{d}^{5}$ - configuration contains maximum number of unpaired electrons i.e. (5). Thus, show maximum dipole moment as per the following relation. Magnetic moment $\left(\mu_{\text {eff }}\right)=\sqrt{n(n+2)} B M$ where, $\mathrm{n}=$ no. of unpaired electrons.
JCECE - 2016
COORDINATION COMPOUNDS
274091
AB is an ionic solid. If the ratio of ionic radii of $A^{+}$and $B^{-}$is 0.52 , what is the coordination number of $B^{-}$?
1 2
2 3
3 6
4 8
Explanation:
(C) : For the ionic solids, we can write the following radius ratio's of the ions which is given below- | Radius ratio $\mathbf{r}^{+} / \mathbf{r}^{-}$ | Co-ordination number | | :--- | :--- | | $0.155-0.225$ | 3 | | $0.225-0.414$ | 4 | | $0.414-0.732$ | 6 | | $0.732-1.00$ | 8 | Hence, if the ratio of ionic radii is 0.52 , then the coordination number of $\mathrm{B}^{-}$is 6 .
2008]**
COORDINATION COMPOUNDS
274070
Which of the following metal ions has a calculated magnetic moment value of $\sqrt{24}$ BM?
274067
The spin only magnetic moment $\left(\mu_{s}\right)$ of a complex $\left[\mathrm{MnBr}_{4}\right]^{2-}$ is $5.9 \mathrm{BM}$. The geometry of the complex will be
1 tetrahedral
2 square planar
3 square pyramidal
4 tetragonal
Explanation:
(A) : In $\left[\mathrm{MnBr}_{4}\right]^{2-}, \mathrm{Mn}$ is present in +2 oxidation state. Number of unpaired electron $=5$ Spin only magnetic moment $=\sqrt{5(5+2)}=$ 5.9 B.M. $\mathrm{Br}^{-}$act as weak field ligand so, pairing of d-electrons not possible. Hence, $\left[\mathrm{MnBr}_{4}\right]^{2-}$ forms outer-orbital complex with $\mathrm{sp}^{3}$ hybridisation and geometry is tetrahedral
AMU-2016
COORDINATION COMPOUNDS
274068
The highest magnetic moment is shown by the transition metal ion with outer electronic configuration
1 $d^{9}$
2 $d^{7}$
3 $\mathrm{d}^{5}$
4 $d^{3}$
Explanation:
(C) : More be the number of unpaired electrons, more be the value of magnetic moment. $\mathrm{d}^{5}$ - configuration contains maximum number of unpaired electrons i.e. (5). Thus, show maximum dipole moment as per the following relation. Magnetic moment $\left(\mu_{\text {eff }}\right)=\sqrt{n(n+2)} B M$ where, $\mathrm{n}=$ no. of unpaired electrons.
JCECE - 2016
COORDINATION COMPOUNDS
274091
AB is an ionic solid. If the ratio of ionic radii of $A^{+}$and $B^{-}$is 0.52 , what is the coordination number of $B^{-}$?
1 2
2 3
3 6
4 8
Explanation:
(C) : For the ionic solids, we can write the following radius ratio's of the ions which is given below- | Radius ratio $\mathbf{r}^{+} / \mathbf{r}^{-}$ | Co-ordination number | | :--- | :--- | | $0.155-0.225$ | 3 | | $0.225-0.414$ | 4 | | $0.414-0.732$ | 6 | | $0.732-1.00$ | 8 | Hence, if the ratio of ionic radii is 0.52 , then the coordination number of $\mathrm{B}^{-}$is 6 .
2008]**
COORDINATION COMPOUNDS
274070
Which of the following metal ions has a calculated magnetic moment value of $\sqrt{24}$ BM?
