(D) : Ferric ion forms a prussian blue precipitate due to the formation of $\mathrm{Fe}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]_{3}$. This complex is nitrogen in the given sample. In this method, to portion of sodium fusion extract, freshly prepared ferrous sulphate, $\mathrm{FeSO}_{4}$ solution is added and warmed. Then about 2 to 3 drops of $\mathrm{FeCl}_{3}$ solution are added and acidified with conc. $\mathrm{HCl}$. The appearance of a Prussian blue colour indicate the presence of nitrogen. $\mathrm{FeSO}_{4}+2 \mathrm{NaOH} \rightarrow \mathrm{Fe}(\mathrm{OH})_{2}+\mathrm{Na}_{2} \mathrm{SO}_{4}$ $6 \mathrm{NaCN}+\mathrm{Fe}(\mathrm{OH})_{2} \rightarrow \underset{\text { Sodium ferrocyanide }}{\mathrm{Na}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]+2 \mathrm{NaOH}}$ $3 \mathrm{Na}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]+4 \mathrm{FeCl}_{3} \rightarrow \underset{\text { Ferric ferrocyanide (prussian blue) }}{\mathrm{Fe}}{ }_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]_{3}+12 \mathrm{NaCl}$
MPPET-2013
COORDINATION COMPOUNDS
274393
Which of the following compounds is not coloured?
(C) : We know that complex compound having no unpaired electron is colourless. Among the given complexes, $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ has no unpaired electron as $\mathrm{CN}^{-}$is a strong field lignad and causes pairing of electrons. So, it is colourless.
VITEEE- 2010
COORDINATION COMPOUNDS
274394
The purple colour of $\mathrm{KMnO}_{4}$ is due to the transition
1 C.T. $(\mathrm{L} \rightarrow \mathrm{M})$
2 C.T.(L $\rightarrow \mathrm{M})$
3 $\mathrm{d}-\mathrm{d}$
4 $\mathrm{p}-\mathrm{d}$
Explanation:
(A) : The permanganate ion has an intense purple colour. Mn (+ VII) has a $\mathrm{d}^{0}$ configuration. So the colour arises from charge transfer and not from $\mathrm{d}-\mathrm{d}$ spectra. In $\mid \mathrm{MnO}_{4}^{-}$an electron is momentarily changing $\mathrm{O}^{--}$to $\mathrm{O}^{-}$ and reducing the oxidation state of the metal from $\mathrm{Mn}(\mathrm{VII})$ to $\mathrm{Mn}(\mathrm{VI})$. Charge transfer requires that the energy levels on the two different atoms are fairly close. $\mathrm{O}=(8)=2_{\mathrm{K}}, 6_{\mathrm{L}}$ $\operatorname{Mn}(25)=2_{\mathrm{K}}, 8_{\mathrm{L}}, 15_{\mathrm{M}}$ Hence, the charge transfer occurs from $\mathrm{L} \rightarrow \mathrm{M}$.
VITEEE- 2007
COORDINATION COMPOUNDS
274396
The orange coloured compound formed when $\mathrm{H}_{2} \mathrm{O}_{2}$ is added to $\mathrm{TiO}_{2}$ solution acidified with conc. $\mathrm{H}_{2} \mathrm{SO}_{4}$ is
1 $\mathrm{Ti}_{2} \mathrm{O}_{3}$
2 $\mathrm{H}_{2} \mathrm{Ti}_{2} \mathrm{O}_{8}$
3 $\mathrm{H}_{2} \mathrm{TiO}_{3}$
4 $\mathrm{H}_{2} \mathrm{TiO}_{4}$
Explanation:
(D) : When an acidified solution of $\mathrm{TiO}_{2}$ is reacted with $\mathrm{H}_{2} \mathrm{O}_{2}$, an intense yellow orange color is obtained due to the formation of $\mathrm{H}_{2} \mathrm{TiO}_{4}$. $\mathrm{TiO}_{2}+\mathrm{H}_{2} \mathrm{O}_{2} \stackrel{\mathrm{H}_{2} \mathrm{SO}_{4}}{\longrightarrow} \mathrm{H}_{2} \mathrm{TiO}_{4}$ $\text { Yellow-orange }$
AP EAMCET- (Engg.) - 2010
COORDINATION COMPOUNDS
274397
Which one of the following gives Prussian blue colour?
(D) : Ferric ion forms a prussian blue precipitate due to the formation of $\mathrm{Fe}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]_{3}$. This complex is nitrogen in the given sample. In this method, to portion of sodium fusion extract, freshly prepared ferrous sulphate, $\mathrm{FeSO}_{4}$ solution is added and warmed. Then about 2 to 3 drops of $\mathrm{FeCl}_{3}$ solution are added and acidified with conc. $\mathrm{HCl}$. The appearance of a Prussian blue colour indicate the presence of nitrogen. $\mathrm{FeSO}_{4}+2 \mathrm{NaOH} \rightarrow \mathrm{Fe}(\mathrm{OH})_{2}+\mathrm{Na}_{2} \mathrm{SO}_{4}$ $6 \mathrm{NaCN}+\mathrm{Fe}(\mathrm{OH})_{2} \rightarrow \underset{\text { Sodium ferrocyanide }}{\mathrm{Na}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]+2 \mathrm{NaOH}}$ $3 \mathrm{Na}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]+4 \mathrm{FeCl}_{3} \rightarrow \underset{\text { Ferric ferrocyanide (prussian blue) }}{\mathrm{Fe}}{ }_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]_{3}+12 \mathrm{NaCl}$
MPPET-2013
COORDINATION COMPOUNDS
274393
Which of the following compounds is not coloured?
(C) : We know that complex compound having no unpaired electron is colourless. Among the given complexes, $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ has no unpaired electron as $\mathrm{CN}^{-}$is a strong field lignad and causes pairing of electrons. So, it is colourless.
VITEEE- 2010
COORDINATION COMPOUNDS
274394
The purple colour of $\mathrm{KMnO}_{4}$ is due to the transition
1 C.T. $(\mathrm{L} \rightarrow \mathrm{M})$
2 C.T.(L $\rightarrow \mathrm{M})$
3 $\mathrm{d}-\mathrm{d}$
4 $\mathrm{p}-\mathrm{d}$
Explanation:
(A) : The permanganate ion has an intense purple colour. Mn (+ VII) has a $\mathrm{d}^{0}$ configuration. So the colour arises from charge transfer and not from $\mathrm{d}-\mathrm{d}$ spectra. In $\mid \mathrm{MnO}_{4}^{-}$an electron is momentarily changing $\mathrm{O}^{--}$to $\mathrm{O}^{-}$ and reducing the oxidation state of the metal from $\mathrm{Mn}(\mathrm{VII})$ to $\mathrm{Mn}(\mathrm{VI})$. Charge transfer requires that the energy levels on the two different atoms are fairly close. $\mathrm{O}=(8)=2_{\mathrm{K}}, 6_{\mathrm{L}}$ $\operatorname{Mn}(25)=2_{\mathrm{K}}, 8_{\mathrm{L}}, 15_{\mathrm{M}}$ Hence, the charge transfer occurs from $\mathrm{L} \rightarrow \mathrm{M}$.
VITEEE- 2007
COORDINATION COMPOUNDS
274396
The orange coloured compound formed when $\mathrm{H}_{2} \mathrm{O}_{2}$ is added to $\mathrm{TiO}_{2}$ solution acidified with conc. $\mathrm{H}_{2} \mathrm{SO}_{4}$ is
1 $\mathrm{Ti}_{2} \mathrm{O}_{3}$
2 $\mathrm{H}_{2} \mathrm{Ti}_{2} \mathrm{O}_{8}$
3 $\mathrm{H}_{2} \mathrm{TiO}_{3}$
4 $\mathrm{H}_{2} \mathrm{TiO}_{4}$
Explanation:
(D) : When an acidified solution of $\mathrm{TiO}_{2}$ is reacted with $\mathrm{H}_{2} \mathrm{O}_{2}$, an intense yellow orange color is obtained due to the formation of $\mathrm{H}_{2} \mathrm{TiO}_{4}$. $\mathrm{TiO}_{2}+\mathrm{H}_{2} \mathrm{O}_{2} \stackrel{\mathrm{H}_{2} \mathrm{SO}_{4}}{\longrightarrow} \mathrm{H}_{2} \mathrm{TiO}_{4}$ $\text { Yellow-orange }$
AP EAMCET- (Engg.) - 2010
COORDINATION COMPOUNDS
274397
Which one of the following gives Prussian blue colour?
(D) : Ferric ion forms a prussian blue precipitate due to the formation of $\mathrm{Fe}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]_{3}$. This complex is nitrogen in the given sample. In this method, to portion of sodium fusion extract, freshly prepared ferrous sulphate, $\mathrm{FeSO}_{4}$ solution is added and warmed. Then about 2 to 3 drops of $\mathrm{FeCl}_{3}$ solution are added and acidified with conc. $\mathrm{HCl}$. The appearance of a Prussian blue colour indicate the presence of nitrogen. $\mathrm{FeSO}_{4}+2 \mathrm{NaOH} \rightarrow \mathrm{Fe}(\mathrm{OH})_{2}+\mathrm{Na}_{2} \mathrm{SO}_{4}$ $6 \mathrm{NaCN}+\mathrm{Fe}(\mathrm{OH})_{2} \rightarrow \underset{\text { Sodium ferrocyanide }}{\mathrm{Na}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]+2 \mathrm{NaOH}}$ $3 \mathrm{Na}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]+4 \mathrm{FeCl}_{3} \rightarrow \underset{\text { Ferric ferrocyanide (prussian blue) }}{\mathrm{Fe}}{ }_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]_{3}+12 \mathrm{NaCl}$
MPPET-2013
COORDINATION COMPOUNDS
274393
Which of the following compounds is not coloured?
(C) : We know that complex compound having no unpaired electron is colourless. Among the given complexes, $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ has no unpaired electron as $\mathrm{CN}^{-}$is a strong field lignad and causes pairing of electrons. So, it is colourless.
VITEEE- 2010
COORDINATION COMPOUNDS
274394
The purple colour of $\mathrm{KMnO}_{4}$ is due to the transition
1 C.T. $(\mathrm{L} \rightarrow \mathrm{M})$
2 C.T.(L $\rightarrow \mathrm{M})$
3 $\mathrm{d}-\mathrm{d}$
4 $\mathrm{p}-\mathrm{d}$
Explanation:
(A) : The permanganate ion has an intense purple colour. Mn (+ VII) has a $\mathrm{d}^{0}$ configuration. So the colour arises from charge transfer and not from $\mathrm{d}-\mathrm{d}$ spectra. In $\mid \mathrm{MnO}_{4}^{-}$an electron is momentarily changing $\mathrm{O}^{--}$to $\mathrm{O}^{-}$ and reducing the oxidation state of the metal from $\mathrm{Mn}(\mathrm{VII})$ to $\mathrm{Mn}(\mathrm{VI})$. Charge transfer requires that the energy levels on the two different atoms are fairly close. $\mathrm{O}=(8)=2_{\mathrm{K}}, 6_{\mathrm{L}}$ $\operatorname{Mn}(25)=2_{\mathrm{K}}, 8_{\mathrm{L}}, 15_{\mathrm{M}}$ Hence, the charge transfer occurs from $\mathrm{L} \rightarrow \mathrm{M}$.
VITEEE- 2007
COORDINATION COMPOUNDS
274396
The orange coloured compound formed when $\mathrm{H}_{2} \mathrm{O}_{2}$ is added to $\mathrm{TiO}_{2}$ solution acidified with conc. $\mathrm{H}_{2} \mathrm{SO}_{4}$ is
1 $\mathrm{Ti}_{2} \mathrm{O}_{3}$
2 $\mathrm{H}_{2} \mathrm{Ti}_{2} \mathrm{O}_{8}$
3 $\mathrm{H}_{2} \mathrm{TiO}_{3}$
4 $\mathrm{H}_{2} \mathrm{TiO}_{4}$
Explanation:
(D) : When an acidified solution of $\mathrm{TiO}_{2}$ is reacted with $\mathrm{H}_{2} \mathrm{O}_{2}$, an intense yellow orange color is obtained due to the formation of $\mathrm{H}_{2} \mathrm{TiO}_{4}$. $\mathrm{TiO}_{2}+\mathrm{H}_{2} \mathrm{O}_{2} \stackrel{\mathrm{H}_{2} \mathrm{SO}_{4}}{\longrightarrow} \mathrm{H}_{2} \mathrm{TiO}_{4}$ $\text { Yellow-orange }$
AP EAMCET- (Engg.) - 2010
COORDINATION COMPOUNDS
274397
Which one of the following gives Prussian blue colour?
(D) : Ferric ion forms a prussian blue precipitate due to the formation of $\mathrm{Fe}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]_{3}$. This complex is nitrogen in the given sample. In this method, to portion of sodium fusion extract, freshly prepared ferrous sulphate, $\mathrm{FeSO}_{4}$ solution is added and warmed. Then about 2 to 3 drops of $\mathrm{FeCl}_{3}$ solution are added and acidified with conc. $\mathrm{HCl}$. The appearance of a Prussian blue colour indicate the presence of nitrogen. $\mathrm{FeSO}_{4}+2 \mathrm{NaOH} \rightarrow \mathrm{Fe}(\mathrm{OH})_{2}+\mathrm{Na}_{2} \mathrm{SO}_{4}$ $6 \mathrm{NaCN}+\mathrm{Fe}(\mathrm{OH})_{2} \rightarrow \underset{\text { Sodium ferrocyanide }}{\mathrm{Na}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]+2 \mathrm{NaOH}}$ $3 \mathrm{Na}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]+4 \mathrm{FeCl}_{3} \rightarrow \underset{\text { Ferric ferrocyanide (prussian blue) }}{\mathrm{Fe}}{ }_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]_{3}+12 \mathrm{NaCl}$
MPPET-2013
COORDINATION COMPOUNDS
274393
Which of the following compounds is not coloured?
(C) : We know that complex compound having no unpaired electron is colourless. Among the given complexes, $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ has no unpaired electron as $\mathrm{CN}^{-}$is a strong field lignad and causes pairing of electrons. So, it is colourless.
VITEEE- 2010
COORDINATION COMPOUNDS
274394
The purple colour of $\mathrm{KMnO}_{4}$ is due to the transition
1 C.T. $(\mathrm{L} \rightarrow \mathrm{M})$
2 C.T.(L $\rightarrow \mathrm{M})$
3 $\mathrm{d}-\mathrm{d}$
4 $\mathrm{p}-\mathrm{d}$
Explanation:
(A) : The permanganate ion has an intense purple colour. Mn (+ VII) has a $\mathrm{d}^{0}$ configuration. So the colour arises from charge transfer and not from $\mathrm{d}-\mathrm{d}$ spectra. In $\mid \mathrm{MnO}_{4}^{-}$an electron is momentarily changing $\mathrm{O}^{--}$to $\mathrm{O}^{-}$ and reducing the oxidation state of the metal from $\mathrm{Mn}(\mathrm{VII})$ to $\mathrm{Mn}(\mathrm{VI})$. Charge transfer requires that the energy levels on the two different atoms are fairly close. $\mathrm{O}=(8)=2_{\mathrm{K}}, 6_{\mathrm{L}}$ $\operatorname{Mn}(25)=2_{\mathrm{K}}, 8_{\mathrm{L}}, 15_{\mathrm{M}}$ Hence, the charge transfer occurs from $\mathrm{L} \rightarrow \mathrm{M}$.
VITEEE- 2007
COORDINATION COMPOUNDS
274396
The orange coloured compound formed when $\mathrm{H}_{2} \mathrm{O}_{2}$ is added to $\mathrm{TiO}_{2}$ solution acidified with conc. $\mathrm{H}_{2} \mathrm{SO}_{4}$ is
1 $\mathrm{Ti}_{2} \mathrm{O}_{3}$
2 $\mathrm{H}_{2} \mathrm{Ti}_{2} \mathrm{O}_{8}$
3 $\mathrm{H}_{2} \mathrm{TiO}_{3}$
4 $\mathrm{H}_{2} \mathrm{TiO}_{4}$
Explanation:
(D) : When an acidified solution of $\mathrm{TiO}_{2}$ is reacted with $\mathrm{H}_{2} \mathrm{O}_{2}$, an intense yellow orange color is obtained due to the formation of $\mathrm{H}_{2} \mathrm{TiO}_{4}$. $\mathrm{TiO}_{2}+\mathrm{H}_{2} \mathrm{O}_{2} \stackrel{\mathrm{H}_{2} \mathrm{SO}_{4}}{\longrightarrow} \mathrm{H}_{2} \mathrm{TiO}_{4}$ $\text { Yellow-orange }$
AP EAMCET- (Engg.) - 2010
COORDINATION COMPOUNDS
274397
Which one of the following gives Prussian blue colour?
(D) : Ferric ion forms a prussian blue precipitate due to the formation of $\mathrm{Fe}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]_{3}$. This complex is nitrogen in the given sample. In this method, to portion of sodium fusion extract, freshly prepared ferrous sulphate, $\mathrm{FeSO}_{4}$ solution is added and warmed. Then about 2 to 3 drops of $\mathrm{FeCl}_{3}$ solution are added and acidified with conc. $\mathrm{HCl}$. The appearance of a Prussian blue colour indicate the presence of nitrogen. $\mathrm{FeSO}_{4}+2 \mathrm{NaOH} \rightarrow \mathrm{Fe}(\mathrm{OH})_{2}+\mathrm{Na}_{2} \mathrm{SO}_{4}$ $6 \mathrm{NaCN}+\mathrm{Fe}(\mathrm{OH})_{2} \rightarrow \underset{\text { Sodium ferrocyanide }}{\mathrm{Na}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]+2 \mathrm{NaOH}}$ $3 \mathrm{Na}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]+4 \mathrm{FeCl}_{3} \rightarrow \underset{\text { Ferric ferrocyanide (prussian blue) }}{\mathrm{Fe}}{ }_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]_{3}+12 \mathrm{NaCl}$
MPPET-2013
COORDINATION COMPOUNDS
274393
Which of the following compounds is not coloured?
(C) : We know that complex compound having no unpaired electron is colourless. Among the given complexes, $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ has no unpaired electron as $\mathrm{CN}^{-}$is a strong field lignad and causes pairing of electrons. So, it is colourless.
VITEEE- 2010
COORDINATION COMPOUNDS
274394
The purple colour of $\mathrm{KMnO}_{4}$ is due to the transition
1 C.T. $(\mathrm{L} \rightarrow \mathrm{M})$
2 C.T.(L $\rightarrow \mathrm{M})$
3 $\mathrm{d}-\mathrm{d}$
4 $\mathrm{p}-\mathrm{d}$
Explanation:
(A) : The permanganate ion has an intense purple colour. Mn (+ VII) has a $\mathrm{d}^{0}$ configuration. So the colour arises from charge transfer and not from $\mathrm{d}-\mathrm{d}$ spectra. In $\mid \mathrm{MnO}_{4}^{-}$an electron is momentarily changing $\mathrm{O}^{--}$to $\mathrm{O}^{-}$ and reducing the oxidation state of the metal from $\mathrm{Mn}(\mathrm{VII})$ to $\mathrm{Mn}(\mathrm{VI})$. Charge transfer requires that the energy levels on the two different atoms are fairly close. $\mathrm{O}=(8)=2_{\mathrm{K}}, 6_{\mathrm{L}}$ $\operatorname{Mn}(25)=2_{\mathrm{K}}, 8_{\mathrm{L}}, 15_{\mathrm{M}}$ Hence, the charge transfer occurs from $\mathrm{L} \rightarrow \mathrm{M}$.
VITEEE- 2007
COORDINATION COMPOUNDS
274396
The orange coloured compound formed when $\mathrm{H}_{2} \mathrm{O}_{2}$ is added to $\mathrm{TiO}_{2}$ solution acidified with conc. $\mathrm{H}_{2} \mathrm{SO}_{4}$ is
1 $\mathrm{Ti}_{2} \mathrm{O}_{3}$
2 $\mathrm{H}_{2} \mathrm{Ti}_{2} \mathrm{O}_{8}$
3 $\mathrm{H}_{2} \mathrm{TiO}_{3}$
4 $\mathrm{H}_{2} \mathrm{TiO}_{4}$
Explanation:
(D) : When an acidified solution of $\mathrm{TiO}_{2}$ is reacted with $\mathrm{H}_{2} \mathrm{O}_{2}$, an intense yellow orange color is obtained due to the formation of $\mathrm{H}_{2} \mathrm{TiO}_{4}$. $\mathrm{TiO}_{2}+\mathrm{H}_{2} \mathrm{O}_{2} \stackrel{\mathrm{H}_{2} \mathrm{SO}_{4}}{\longrightarrow} \mathrm{H}_{2} \mathrm{TiO}_{4}$ $\text { Yellow-orange }$
AP EAMCET- (Engg.) - 2010
COORDINATION COMPOUNDS
274397
Which one of the following gives Prussian blue colour?
