The Crystal Field Stabilization Energy (CFSE) and the spin-only magnetic moment in Bohr Magneton (BM) for the complex ${\mathrm{K}}_3\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]$ are, respectively,

1.  $0.0{∆}_0 \mathrm{and} \sqrt{35} \mathrm{BM}$

2.  $-2.0{∆}_0 \mathrm{and} \sqrt{3} \mathrm{BM}$

3.  $-0.4{∆}_0 \mathrm{and} \sqrt{24} \mathrm{BM}$

4.  $-2.4{∆}_0 \mathrm{and} \sqrt{0} \mathrm{BM}$

${\mathrm{Zn}}^{2+} + 4{\mathrm{NH}}_3 \rightleftharpoons {\left[\mathrm{Zn}{\left({\mathrm{NH}}_3\right)}_4\right]}^{2+}$. The concentration of free ${\mathrm{Zn}}^{2+}$ ions in a solution of the complex ion ${\left[\mathrm{Zn}{\left({\mathrm{NH}}_3\right)}_4\right]}^{2+}$ can be increased significantly by the addition of

1.  ${\mathrm{H}}_2\mathrm{O}$

2.  HCl(aq)

3.  ${\mathrm{NH}}_3\left(\mathrm{aq}\right)$

4.  ${\mathrm{NH}}_4\mathrm{Cl}\left(\mathrm{aq}\right)$

Given the following pair of isomers:
[Cr(NH₃)₆][Cr(NO₂)₆] and
[Cr(NH₃)₄(NO₂)₂][Cr(NH₃)₂(NO₂)₄]
Aside from X-ray diffraction, how could the following pairs of isomers be distinguished from one another?

1. Cryoscopic method
2. Measuring their molar conductance
3. Measuring their magnetic moments
4. Observing their colors

The hybridizations of $\left[\mathrm{Ni}{\left(\mathrm{CO}\right)}_4\right] \mathrm{and} {\left[\mathrm{Cr}\left({\mathrm{H}}_2\mathrm{O}{)}_6\right)\right]}^{2+}$, respectively, are : 

1.  ${\mathrm{sp}}^3 \mathrm{and} {\mathrm{d}}^3{\mathrm{sp}}^2$

2.  ${\mathrm{dsp}}^2 \mathrm{and} {\mathrm{d}}^2{\mathrm{sp}}^3$

3.  ${\mathrm{sp}}^3 \mathrm{and} {\mathrm{d}}^2{\mathrm{sp}}^3$

4.  ${\mathrm{dsp}}^2 \mathrm{and} {\mathrm{sp}}^3{\mathrm{d}}^2$

Which of the following is true about the complex $\left[{\mathrm{PtCl}}_2\left({\mathrm{NH}}_3\right) \left(H20\right)\right]$; [Atomic number of Pt = 78]

I.  It will have two geometrical isomeric forms, cis and trans

II.  The hybridization state of Pt(ll) is ${\mathrm{sp}}^3$

III.  It is a square planar complex

IV.  It is a diamagnetic complex

V.  It can show hydrate isomerism

VI.  It is a tetrahedral complex

1.  I, III, IV

2.  II, IV, V

3.  II, V, IV

4.  I, V, IV

Correct statement among the following regarding $\mathrm{Ni}{\left(\mathrm{CO}\right)}_4$, ${\left[\mathrm{Ni}{\left(\mathrm{CN}\right)}_4\right]}^{2-}$ and ${\left[{\mathrm{NiCl}}_4\right]}^{2-}$ is -

1.  $\mathrm{Ni}{\left(\mathrm{CO}\right)}_4$ and ${\left[{\mathrm{NiCl}}_4\right]}^{2-}$ are diamagnetic and ${\left[\mathrm{Ni}{\left(\mathrm{CN}\right)}_4\right]}^{2-}$ is paramagnetic

2.  ${\left[{\mathrm{NiCl}}_4\right]}^{2-}$ and ${\left[\mathrm{Ni}{\left(\mathrm{CN}\right)}_4\right]}^{2-}$ are diamagnetic and $\mathrm{Ni}{\left(\mathrm{CO}\right)}_4$ is paramagnetic

3.  $\mathrm{Ni}{\left(\mathrm{CO}\right)}_4$ and ${\left[\mathrm{Ni}{\left(\mathrm{CN}\right)}_4\right]}^{2-}$ are diamagnetic and ${\left[{\mathrm{NiCl}}_4\right]}^{2-}$ is paramagnetic

4.  $\mathrm{Ni}{\left(\mathrm{CO}\right)}_4$ is diamagnetic and ${\left[{\mathrm{NiCl}}_4\right]}^{2-}$ and ${\left[\mathrm{Ni}{\left(\mathrm{CN}\right)}_4\right]}^{2-}$ are paramagnetic

Among the following statements, the correct statement is/are -

a.  ${\left[\mathrm{Ni}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{2+}$ and ${\left[\mathrm{Ni}{\left({\mathrm{NH}}_3\right)}_6\right]}^{2+}$ have the same value of CFSE

b.  ${\left[\mathrm{Ni}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{2+}$ and ${\left[\mathrm{Ni}{\left({\mathrm{NH}}_3\right)}_6\right]}^{2+}$ have the same value of the magnetic moment

c.  ${\left[\mathrm{Ni}{\left({\mathrm{H}}_2\mathrm{O}\right)}_{6^{}}\right]}^{2+}$ and ${\left[\mathrm{Ni}{\left({\mathrm{NH}}_3\right)}_6\right]}^{2+}$ have the same hybridization of nickel

1.  a & c only

2.  b & c only

3.  a & b only

4. All of the above

Which is the correct electronic configuration for the complex ${\left[\mathrm{Ni}{\left({\mathrm{NH}}_3\right)}_6\right]}^{2+}$?

1.  ${\mathrm{t}}_{2\mathrm{g}}^4 {\mathrm{e}}_{\mathrm{g}}^2$

2.  ${\mathrm{t}}_{2\mathrm{g}}^6 {\mathrm{e}}_{\mathrm{g}}^2$

3.  ${\mathrm{t}}_{2\mathrm{g}}^7 {\mathrm{e}}_{\mathrm{g}}^1$

4.  ${\mathrm{t}}_{2\mathrm{g}}^4 {\mathrm{e}}_{\mathrm{g}}^4$

A compound is made by mixing cobalt (III) nitrate and potassium nitrite solution in the ratio of 1 : 3. The aqueous solution of the compound showed 4 particles per molecule whereas molar conductivity reveals the presence of six electrical. charges. The formula of the compound is

1.  $\mathrm{Co}{\left({\mathrm{NO}}_2\right)}_3 · 2{\mathrm{KNO}}_2$

2.  $\mathrm{Co}{\left({\mathrm{NO}}_2\right)}_3 · 3{\mathrm{KNO}}_2$

3.  ${\mathrm{K}}_3\left[\mathrm{Co}{\left({\mathrm{NO}}_2\right)}_6\right]$

4.  $\mathrm{K}\left[\mathrm{Co}{\left({\mathrm{NO}}_2\right)}_4\right]$

0.001 mole of $[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_5\left({\mathrm{NO}}_3\right) \left({\mathrm{SO}}_4\right)]$ was passed through a cation exchanger and the acid coming out of it required 20 ml of 0.1 M NaOH for neutralization. Hence, the complex is

1.  $\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_5{\mathrm{SO}}_4\right]{\mathrm{NO}}_3$

2.  $\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_5 N{O}_3\right]{\mathrm{SO}}_4$

3.  $\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_5{\mathrm{SO}}_4\right]{\mathrm{NO}}_3{\mathrm{SO}}_4$

4.  All of these