274299
What is the correct electronic configuration of the central atom in $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ based on crystal field theory?
(C) : $\mathrm{K}_{4}\left(\mathrm{Fe}(\mathrm{CN})_{6}\right)$ Let, $\mathrm{X}$ be the oxidation state of $\mathrm{fe}$ $\therefore 4(+1)+\mathrm{x}+6(-1)=0$ $\mathrm{Fe}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^{6} \mathrm{x}=+2$ Pairing occure because $\mathrm{CN}$ is strong field ligand (SFL) Here, $\mathrm{t}_{2 \mathrm{~g}}^{6} \mathrm{e}_{\mathrm{g}}^{0}$ is the correct configuration $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$.
NEET 2019
COORDINATION COMPOUNDS
274302
$\left[\mathrm{Co}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]^{3-}$ is a:
1 Low spin complex
2 Paramagnetic
3 High Spin
4 $\mathrm{sp}^{3} \mathrm{~d}^{2}$ hybridized
Explanation:
(A) : $\left[\mathrm{Co}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]^{3-}$ Let, $\mathrm{x}$ be the oxidation state of $\mathrm{Co}$. $\mathrm{x}=3(-2)=-3$ $\mathrm{x}=+3$ $\mathrm{C}_{2} \mathrm{O}_{4}$ is a bidentate ligand and this ligand form the octahedral complex. The complex exhibit the low spin complex because of bidentate nature of the ligand that's why pairing of electron occure in the inner orbital. $\mathrm{Co}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^{6}-\mathrm{v}_{\mathrm{g}} \uparrow \downarrow \downarrow_{\mathrm{t}_{2 \mathrm{~g}}}$ Hence, it form the low spin complex and it is diamagnetic in nature. The hybridisation of the compound is $\mathrm{d}^{2} \mathrm{sp}^{3}$.
AIIMS 25 May 2019 (Evening)
COORDINATION COMPOUNDS
274303
Which of the following complexes has maximum CFSE?
(D) : The crystal field stabilisation energy depends upon the following factor- (1) Charge on metal, (2) Size of metal (3) Nature of ligand Here, charge of metal and nature of ligand are same i.e. +3 and $\mathrm{NH}_{3}$. The unsimilarities arises in the size of metal, as we know that as well as size of metal incrases, the CFSE of complex is also increases. Hence, the correct order or CFSE will be- $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}<\left[\mathrm{Rh}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}<\left[\mathrm{Ir}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}$
274299
What is the correct electronic configuration of the central atom in $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ based on crystal field theory?
(C) : $\mathrm{K}_{4}\left(\mathrm{Fe}(\mathrm{CN})_{6}\right)$ Let, $\mathrm{X}$ be the oxidation state of $\mathrm{fe}$ $\therefore 4(+1)+\mathrm{x}+6(-1)=0$ $\mathrm{Fe}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^{6} \mathrm{x}=+2$ Pairing occure because $\mathrm{CN}$ is strong field ligand (SFL) Here, $\mathrm{t}_{2 \mathrm{~g}}^{6} \mathrm{e}_{\mathrm{g}}^{0}$ is the correct configuration $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$.
NEET 2019
COORDINATION COMPOUNDS
274302
$\left[\mathrm{Co}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]^{3-}$ is a:
1 Low spin complex
2 Paramagnetic
3 High Spin
4 $\mathrm{sp}^{3} \mathrm{~d}^{2}$ hybridized
Explanation:
(A) : $\left[\mathrm{Co}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]^{3-}$ Let, $\mathrm{x}$ be the oxidation state of $\mathrm{Co}$. $\mathrm{x}=3(-2)=-3$ $\mathrm{x}=+3$ $\mathrm{C}_{2} \mathrm{O}_{4}$ is a bidentate ligand and this ligand form the octahedral complex. The complex exhibit the low spin complex because of bidentate nature of the ligand that's why pairing of electron occure in the inner orbital. $\mathrm{Co}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^{6}-\mathrm{v}_{\mathrm{g}} \uparrow \downarrow \downarrow_{\mathrm{t}_{2 \mathrm{~g}}}$ Hence, it form the low spin complex and it is diamagnetic in nature. The hybridisation of the compound is $\mathrm{d}^{2} \mathrm{sp}^{3}$.
AIIMS 25 May 2019 (Evening)
COORDINATION COMPOUNDS
274303
Which of the following complexes has maximum CFSE?
(D) : The crystal field stabilisation energy depends upon the following factor- (1) Charge on metal, (2) Size of metal (3) Nature of ligand Here, charge of metal and nature of ligand are same i.e. +3 and $\mathrm{NH}_{3}$. The unsimilarities arises in the size of metal, as we know that as well as size of metal incrases, the CFSE of complex is also increases. Hence, the correct order or CFSE will be- $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}<\left[\mathrm{Rh}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}<\left[\mathrm{Ir}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}$
274299
What is the correct electronic configuration of the central atom in $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ based on crystal field theory?
(C) : $\mathrm{K}_{4}\left(\mathrm{Fe}(\mathrm{CN})_{6}\right)$ Let, $\mathrm{X}$ be the oxidation state of $\mathrm{fe}$ $\therefore 4(+1)+\mathrm{x}+6(-1)=0$ $\mathrm{Fe}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^{6} \mathrm{x}=+2$ Pairing occure because $\mathrm{CN}$ is strong field ligand (SFL) Here, $\mathrm{t}_{2 \mathrm{~g}}^{6} \mathrm{e}_{\mathrm{g}}^{0}$ is the correct configuration $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$.
NEET 2019
COORDINATION COMPOUNDS
274302
$\left[\mathrm{Co}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]^{3-}$ is a:
1 Low spin complex
2 Paramagnetic
3 High Spin
4 $\mathrm{sp}^{3} \mathrm{~d}^{2}$ hybridized
Explanation:
(A) : $\left[\mathrm{Co}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]^{3-}$ Let, $\mathrm{x}$ be the oxidation state of $\mathrm{Co}$. $\mathrm{x}=3(-2)=-3$ $\mathrm{x}=+3$ $\mathrm{C}_{2} \mathrm{O}_{4}$ is a bidentate ligand and this ligand form the octahedral complex. The complex exhibit the low spin complex because of bidentate nature of the ligand that's why pairing of electron occure in the inner orbital. $\mathrm{Co}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^{6}-\mathrm{v}_{\mathrm{g}} \uparrow \downarrow \downarrow_{\mathrm{t}_{2 \mathrm{~g}}}$ Hence, it form the low spin complex and it is diamagnetic in nature. The hybridisation of the compound is $\mathrm{d}^{2} \mathrm{sp}^{3}$.
AIIMS 25 May 2019 (Evening)
COORDINATION COMPOUNDS
274303
Which of the following complexes has maximum CFSE?
(D) : The crystal field stabilisation energy depends upon the following factor- (1) Charge on metal, (2) Size of metal (3) Nature of ligand Here, charge of metal and nature of ligand are same i.e. +3 and $\mathrm{NH}_{3}$. The unsimilarities arises in the size of metal, as we know that as well as size of metal incrases, the CFSE of complex is also increases. Hence, the correct order or CFSE will be- $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}<\left[\mathrm{Rh}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}<\left[\mathrm{Ir}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}$
274299
What is the correct electronic configuration of the central atom in $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ based on crystal field theory?
(C) : $\mathrm{K}_{4}\left(\mathrm{Fe}(\mathrm{CN})_{6}\right)$ Let, $\mathrm{X}$ be the oxidation state of $\mathrm{fe}$ $\therefore 4(+1)+\mathrm{x}+6(-1)=0$ $\mathrm{Fe}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^{6} \mathrm{x}=+2$ Pairing occure because $\mathrm{CN}$ is strong field ligand (SFL) Here, $\mathrm{t}_{2 \mathrm{~g}}^{6} \mathrm{e}_{\mathrm{g}}^{0}$ is the correct configuration $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$.
NEET 2019
COORDINATION COMPOUNDS
274302
$\left[\mathrm{Co}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]^{3-}$ is a:
1 Low spin complex
2 Paramagnetic
3 High Spin
4 $\mathrm{sp}^{3} \mathrm{~d}^{2}$ hybridized
Explanation:
(A) : $\left[\mathrm{Co}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]^{3-}$ Let, $\mathrm{x}$ be the oxidation state of $\mathrm{Co}$. $\mathrm{x}=3(-2)=-3$ $\mathrm{x}=+3$ $\mathrm{C}_{2} \mathrm{O}_{4}$ is a bidentate ligand and this ligand form the octahedral complex. The complex exhibit the low spin complex because of bidentate nature of the ligand that's why pairing of electron occure in the inner orbital. $\mathrm{Co}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^{6}-\mathrm{v}_{\mathrm{g}} \uparrow \downarrow \downarrow_{\mathrm{t}_{2 \mathrm{~g}}}$ Hence, it form the low spin complex and it is diamagnetic in nature. The hybridisation of the compound is $\mathrm{d}^{2} \mathrm{sp}^{3}$.
AIIMS 25 May 2019 (Evening)
COORDINATION COMPOUNDS
274303
Which of the following complexes has maximum CFSE?
(D) : The crystal field stabilisation energy depends upon the following factor- (1) Charge on metal, (2) Size of metal (3) Nature of ligand Here, charge of metal and nature of ligand are same i.e. +3 and $\mathrm{NH}_{3}$. The unsimilarities arises in the size of metal, as we know that as well as size of metal incrases, the CFSE of complex is also increases. Hence, the correct order or CFSE will be- $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}<\left[\mathrm{Rh}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}<\left[\mathrm{Ir}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}$
