274251
Which of the following facts about the complex $\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}$ is wrong?
1 The complex involves $\mathrm{d}^{2} \mathrm{sp}^{3}$ hybridization and is octahedral in shape
2 The complex is paramagnetic
3 The complex is an outer orbital complex
4 The complex gives white precipitate with silver nitrate solution
Explanation:
(C) : $\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}$ complex, $\mathrm{Cr}$ is present in +3 oxidation. In $\mathrm{Cr}^{3+}$ always inner d-orbital i.e., $(\mathrm{n}-1) \mathrm{d}$ is used for hybridisation and forms $\mathrm{d}^{2} \mathrm{sp}^{3}$ hybridisation with octahedral geometry. $\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}$ contains three unpaired electrons therefore it shows paramagnetic behavior. ${\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right] \cdot \mathrm{Cl}_{3}+\mathrm{AgNO}_{3} \longrightarrow}$ $\text { (ppt.) } 3$
AIEEE 2011
COORDINATION COMPOUNDS
274253
The effective atomic number for $\left[\mathrm{Rh}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}$ (atomic number of $R h$ is 45 ) is
1 42
2 45
3 48
4 54
Explanation:
(D) : Atomic Number of $\mathrm{Rh}=45$ Oxidation state of $\mathrm{Rh}=+3$ No. of donor site $=6$ EAN of $\left[\mathrm{Rh}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}=$ Atomic no. - Oxidation $+2 \times$ Number of donor atom attached with metal. $=45-3+2 \times 6=54$
J and K CET-(2012)
COORDINATION COMPOUNDS
274257
The metal ion in complex $A$ has EAN identical to the atomic number of krypton $A$ is (At. $\mathrm{No}$ of $\mathrm{Cr}=24, \mathrm{Fe}=26, \mathrm{Pd}=46$ )
(C) : Atomic number of krypton $=36$ E.AN of complex A = Atomic No of Krypton Atomic number of $\mathrm{Fe}=26$ Oxidation state of $\mathrm{Fe}$ in $\mathrm{Na}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ complex $x+6(-1)=-4$ $x=2$ Number of donor sites attached with $\mathrm{Fe}=6$ E.AN of $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}=\mathrm{Z}-(\mathrm{ON})+2(\mathrm{CN})$ Where, $\mathrm{ON}=$ Oxidation no. $\mathrm{CN}=$ coordination number $\mathrm{Z}=$ atomic number $=26-2+2(6)$ $=36$ Hence, complex A is $\mathrm{Na}_{4}[\mathrm{Fe}(\mathrm{CN})]_{6}$
J and K CET-(2006)
COORDINATION COMPOUNDS
274189
What is the value of $x$ on the $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{x}$ complex ion?
1 +2
2 -2
3 0 (zero)
4 +4
Explanation:
(B) : Nickel forms square planar complex with coordination number 4 and strong field ligand with oxidation state is +2 . $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{\mathrm{x}}$ $2+4 \times(-1)=\mathrm{x}$ $2-4=x$ $\mathrm{x}=-2$ The complex is $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$
Manipal-2020
COORDINATION COMPOUNDS
274165
$\mathrm{Fe}^{3+}$ cation gives a prussian blue precipitate on addition of potassium ferrocyanide solution due to the formation of:
274251
Which of the following facts about the complex $\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}$ is wrong?
1 The complex involves $\mathrm{d}^{2} \mathrm{sp}^{3}$ hybridization and is octahedral in shape
2 The complex is paramagnetic
3 The complex is an outer orbital complex
4 The complex gives white precipitate with silver nitrate solution
Explanation:
(C) : $\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}$ complex, $\mathrm{Cr}$ is present in +3 oxidation. In $\mathrm{Cr}^{3+}$ always inner d-orbital i.e., $(\mathrm{n}-1) \mathrm{d}$ is used for hybridisation and forms $\mathrm{d}^{2} \mathrm{sp}^{3}$ hybridisation with octahedral geometry. $\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}$ contains three unpaired electrons therefore it shows paramagnetic behavior. ${\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right] \cdot \mathrm{Cl}_{3}+\mathrm{AgNO}_{3} \longrightarrow}$ $\text { (ppt.) } 3$
AIEEE 2011
COORDINATION COMPOUNDS
274253
The effective atomic number for $\left[\mathrm{Rh}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}$ (atomic number of $R h$ is 45 ) is
1 42
2 45
3 48
4 54
Explanation:
(D) : Atomic Number of $\mathrm{Rh}=45$ Oxidation state of $\mathrm{Rh}=+3$ No. of donor site $=6$ EAN of $\left[\mathrm{Rh}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}=$ Atomic no. - Oxidation $+2 \times$ Number of donor atom attached with metal. $=45-3+2 \times 6=54$
J and K CET-(2012)
COORDINATION COMPOUNDS
274257
The metal ion in complex $A$ has EAN identical to the atomic number of krypton $A$ is (At. $\mathrm{No}$ of $\mathrm{Cr}=24, \mathrm{Fe}=26, \mathrm{Pd}=46$ )
(C) : Atomic number of krypton $=36$ E.AN of complex A = Atomic No of Krypton Atomic number of $\mathrm{Fe}=26$ Oxidation state of $\mathrm{Fe}$ in $\mathrm{Na}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ complex $x+6(-1)=-4$ $x=2$ Number of donor sites attached with $\mathrm{Fe}=6$ E.AN of $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}=\mathrm{Z}-(\mathrm{ON})+2(\mathrm{CN})$ Where, $\mathrm{ON}=$ Oxidation no. $\mathrm{CN}=$ coordination number $\mathrm{Z}=$ atomic number $=26-2+2(6)$ $=36$ Hence, complex A is $\mathrm{Na}_{4}[\mathrm{Fe}(\mathrm{CN})]_{6}$
J and K CET-(2006)
COORDINATION COMPOUNDS
274189
What is the value of $x$ on the $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{x}$ complex ion?
1 +2
2 -2
3 0 (zero)
4 +4
Explanation:
(B) : Nickel forms square planar complex with coordination number 4 and strong field ligand with oxidation state is +2 . $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{\mathrm{x}}$ $2+4 \times(-1)=\mathrm{x}$ $2-4=x$ $\mathrm{x}=-2$ The complex is $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$
Manipal-2020
COORDINATION COMPOUNDS
274165
$\mathrm{Fe}^{3+}$ cation gives a prussian blue precipitate on addition of potassium ferrocyanide solution due to the formation of:
274251
Which of the following facts about the complex $\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}$ is wrong?
1 The complex involves $\mathrm{d}^{2} \mathrm{sp}^{3}$ hybridization and is octahedral in shape
2 The complex is paramagnetic
3 The complex is an outer orbital complex
4 The complex gives white precipitate with silver nitrate solution
Explanation:
(C) : $\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}$ complex, $\mathrm{Cr}$ is present in +3 oxidation. In $\mathrm{Cr}^{3+}$ always inner d-orbital i.e., $(\mathrm{n}-1) \mathrm{d}$ is used for hybridisation and forms $\mathrm{d}^{2} \mathrm{sp}^{3}$ hybridisation with octahedral geometry. $\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}$ contains three unpaired electrons therefore it shows paramagnetic behavior. ${\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right] \cdot \mathrm{Cl}_{3}+\mathrm{AgNO}_{3} \longrightarrow}$ $\text { (ppt.) } 3$
AIEEE 2011
COORDINATION COMPOUNDS
274253
The effective atomic number for $\left[\mathrm{Rh}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}$ (atomic number of $R h$ is 45 ) is
1 42
2 45
3 48
4 54
Explanation:
(D) : Atomic Number of $\mathrm{Rh}=45$ Oxidation state of $\mathrm{Rh}=+3$ No. of donor site $=6$ EAN of $\left[\mathrm{Rh}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}=$ Atomic no. - Oxidation $+2 \times$ Number of donor atom attached with metal. $=45-3+2 \times 6=54$
J and K CET-(2012)
COORDINATION COMPOUNDS
274257
The metal ion in complex $A$ has EAN identical to the atomic number of krypton $A$ is (At. $\mathrm{No}$ of $\mathrm{Cr}=24, \mathrm{Fe}=26, \mathrm{Pd}=46$ )
(C) : Atomic number of krypton $=36$ E.AN of complex A = Atomic No of Krypton Atomic number of $\mathrm{Fe}=26$ Oxidation state of $\mathrm{Fe}$ in $\mathrm{Na}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ complex $x+6(-1)=-4$ $x=2$ Number of donor sites attached with $\mathrm{Fe}=6$ E.AN of $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}=\mathrm{Z}-(\mathrm{ON})+2(\mathrm{CN})$ Where, $\mathrm{ON}=$ Oxidation no. $\mathrm{CN}=$ coordination number $\mathrm{Z}=$ atomic number $=26-2+2(6)$ $=36$ Hence, complex A is $\mathrm{Na}_{4}[\mathrm{Fe}(\mathrm{CN})]_{6}$
J and K CET-(2006)
COORDINATION COMPOUNDS
274189
What is the value of $x$ on the $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{x}$ complex ion?
1 +2
2 -2
3 0 (zero)
4 +4
Explanation:
(B) : Nickel forms square planar complex with coordination number 4 and strong field ligand with oxidation state is +2 . $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{\mathrm{x}}$ $2+4 \times(-1)=\mathrm{x}$ $2-4=x$ $\mathrm{x}=-2$ The complex is $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$
Manipal-2020
COORDINATION COMPOUNDS
274165
$\mathrm{Fe}^{3+}$ cation gives a prussian blue precipitate on addition of potassium ferrocyanide solution due to the formation of:
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
COORDINATION COMPOUNDS
274251
Which of the following facts about the complex $\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}$ is wrong?
1 The complex involves $\mathrm{d}^{2} \mathrm{sp}^{3}$ hybridization and is octahedral in shape
2 The complex is paramagnetic
3 The complex is an outer orbital complex
4 The complex gives white precipitate with silver nitrate solution
Explanation:
(C) : $\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}$ complex, $\mathrm{Cr}$ is present in +3 oxidation. In $\mathrm{Cr}^{3+}$ always inner d-orbital i.e., $(\mathrm{n}-1) \mathrm{d}$ is used for hybridisation and forms $\mathrm{d}^{2} \mathrm{sp}^{3}$ hybridisation with octahedral geometry. $\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}$ contains three unpaired electrons therefore it shows paramagnetic behavior. ${\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right] \cdot \mathrm{Cl}_{3}+\mathrm{AgNO}_{3} \longrightarrow}$ $\text { (ppt.) } 3$
AIEEE 2011
COORDINATION COMPOUNDS
274253
The effective atomic number for $\left[\mathrm{Rh}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}$ (atomic number of $R h$ is 45 ) is
1 42
2 45
3 48
4 54
Explanation:
(D) : Atomic Number of $\mathrm{Rh}=45$ Oxidation state of $\mathrm{Rh}=+3$ No. of donor site $=6$ EAN of $\left[\mathrm{Rh}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}=$ Atomic no. - Oxidation $+2 \times$ Number of donor atom attached with metal. $=45-3+2 \times 6=54$
J and K CET-(2012)
COORDINATION COMPOUNDS
274257
The metal ion in complex $A$ has EAN identical to the atomic number of krypton $A$ is (At. $\mathrm{No}$ of $\mathrm{Cr}=24, \mathrm{Fe}=26, \mathrm{Pd}=46$ )
(C) : Atomic number of krypton $=36$ E.AN of complex A = Atomic No of Krypton Atomic number of $\mathrm{Fe}=26$ Oxidation state of $\mathrm{Fe}$ in $\mathrm{Na}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ complex $x+6(-1)=-4$ $x=2$ Number of donor sites attached with $\mathrm{Fe}=6$ E.AN of $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}=\mathrm{Z}-(\mathrm{ON})+2(\mathrm{CN})$ Where, $\mathrm{ON}=$ Oxidation no. $\mathrm{CN}=$ coordination number $\mathrm{Z}=$ atomic number $=26-2+2(6)$ $=36$ Hence, complex A is $\mathrm{Na}_{4}[\mathrm{Fe}(\mathrm{CN})]_{6}$
J and K CET-(2006)
COORDINATION COMPOUNDS
274189
What is the value of $x$ on the $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{x}$ complex ion?
1 +2
2 -2
3 0 (zero)
4 +4
Explanation:
(B) : Nickel forms square planar complex with coordination number 4 and strong field ligand with oxidation state is +2 . $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{\mathrm{x}}$ $2+4 \times(-1)=\mathrm{x}$ $2-4=x$ $\mathrm{x}=-2$ The complex is $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$
Manipal-2020
COORDINATION COMPOUNDS
274165
$\mathrm{Fe}^{3+}$ cation gives a prussian blue precipitate on addition of potassium ferrocyanide solution due to the formation of:
274251
Which of the following facts about the complex $\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}$ is wrong?
1 The complex involves $\mathrm{d}^{2} \mathrm{sp}^{3}$ hybridization and is octahedral in shape
2 The complex is paramagnetic
3 The complex is an outer orbital complex
4 The complex gives white precipitate with silver nitrate solution
Explanation:
(C) : $\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}$ complex, $\mathrm{Cr}$ is present in +3 oxidation. In $\mathrm{Cr}^{3+}$ always inner d-orbital i.e., $(\mathrm{n}-1) \mathrm{d}$ is used for hybridisation and forms $\mathrm{d}^{2} \mathrm{sp}^{3}$ hybridisation with octahedral geometry. $\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}$ contains three unpaired electrons therefore it shows paramagnetic behavior. ${\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right] \cdot \mathrm{Cl}_{3}+\mathrm{AgNO}_{3} \longrightarrow}$ $\text { (ppt.) } 3$
AIEEE 2011
COORDINATION COMPOUNDS
274253
The effective atomic number for $\left[\mathrm{Rh}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}$ (atomic number of $R h$ is 45 ) is
1 42
2 45
3 48
4 54
Explanation:
(D) : Atomic Number of $\mathrm{Rh}=45$ Oxidation state of $\mathrm{Rh}=+3$ No. of donor site $=6$ EAN of $\left[\mathrm{Rh}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}=$ Atomic no. - Oxidation $+2 \times$ Number of donor atom attached with metal. $=45-3+2 \times 6=54$
J and K CET-(2012)
COORDINATION COMPOUNDS
274257
The metal ion in complex $A$ has EAN identical to the atomic number of krypton $A$ is (At. $\mathrm{No}$ of $\mathrm{Cr}=24, \mathrm{Fe}=26, \mathrm{Pd}=46$ )
(C) : Atomic number of krypton $=36$ E.AN of complex A = Atomic No of Krypton Atomic number of $\mathrm{Fe}=26$ Oxidation state of $\mathrm{Fe}$ in $\mathrm{Na}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ complex $x+6(-1)=-4$ $x=2$ Number of donor sites attached with $\mathrm{Fe}=6$ E.AN of $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}=\mathrm{Z}-(\mathrm{ON})+2(\mathrm{CN})$ Where, $\mathrm{ON}=$ Oxidation no. $\mathrm{CN}=$ coordination number $\mathrm{Z}=$ atomic number $=26-2+2(6)$ $=36$ Hence, complex A is $\mathrm{Na}_{4}[\mathrm{Fe}(\mathrm{CN})]_{6}$
J and K CET-(2006)
COORDINATION COMPOUNDS
274189
What is the value of $x$ on the $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{x}$ complex ion?
1 +2
2 -2
3 0 (zero)
4 +4
Explanation:
(B) : Nickel forms square planar complex with coordination number 4 and strong field ligand with oxidation state is +2 . $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{\mathrm{x}}$ $2+4 \times(-1)=\mathrm{x}$ $2-4=x$ $\mathrm{x}=-2$ The complex is $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$
Manipal-2020
COORDINATION COMPOUNDS
274165
$\mathrm{Fe}^{3+}$ cation gives a prussian blue precipitate on addition of potassium ferrocyanide solution due to the formation of:
