A solution of [Ni(H₂O)₆]²⁺ is green, but a solution of [Ni(CN)₄]^2– is colorless because:

1.	There are paired electrons in [Ni(H2O)6]2+ while all electrons are unpaired in [Ni(CN)4]2–
2.	There are unpaired electrons in [Ni(H2O)6]2+ while all electrons are paired in [Ni(CN)4]2–
3.	There are unpaired electrons in [Ni(H2O)6]2+ and [Ni(CN)4]2–
4.	None of the above.

[Fe(CN)₆]^4– and [Fe(H₂O)₆]²⁺ are of different colors in dilute solutions because :

The nature of bonding in metal carbonyls is -

The greater value of the stability constant of complex salt indicates:

1. The greater proportion of products.

2. The greater proportion of reactants

3. Lesser concentration of catalyst at equilibrium

4. All of the above

The correct order for the wavelengths of absorption in the visible region for the following is:
\(\left[\mathrm{Ni}\left(\mathrm{NO}_2\right)_6\right]^{4-},\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_6\right]^{2+},\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}\)
1. $[\mathrm{Ni}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6{]}^{2+}$ < $[\mathrm{Ni}{\left({\mathrm{NH}}_3\right)}_6{]}^{2+}$ < $[\mathrm{Ni}{\left({\mathrm{NO}}_2\right)}_6{]}^{4-}$

2. $[\mathrm{Ni}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6{]}^{2+}$ > $[\mathrm{Ni}{\left({\mathrm{NH}}_3\right)}_6{]}^{2+}$ > $[\mathrm{Ni}{\left({\mathrm{NO}}_2\right)}_6{]}^{4-}$

3. $[\mathrm{Ni}{\left({\mathrm{NH}}_3\right)}_6{]}^{2+}$ > $[\mathrm{Ni}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6{]}^{2+}$ > $[\mathrm{Ni}{\left({\mathrm{NO}}_2\right)}_6{]}^{4-}$

4.  $[\mathrm{Ni}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6{]}^{2+}$ > ${\left[\mathrm{Ni}{\left({\mathrm{NO}}_2\right)}_6\right]}^{4-}$ > [Ni(NH₃)₆]²⁺

The number of unpaired electrons in the following complexes respectively are-

${\mathrm{K}}_4\left[\mathrm{Mn}{\left(\mathrm{CN}\right)}_6\right], \mathrm{Fe}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6{]}^{2+}, {\mathrm{K}}_2\left[{\mathrm{MnCl}}_4\right]$

1. 1, 4, 5

2. 2, 4, 5

3. 0, 3, 4

4. 2, 3, 4

Match Column-I with Column-II and mark the correct option.

Codes:
1. i=d, ii=c, iii=b, iv=a

2. i=c, ii=d, iii=b, iv=a

3. i=c, ii=d, iii=a, iv=b

4. i=a, ii=c, iii=b, iv=d

The correct representation of tetrahydroxidozincate (II) is:
1. \([Zn(OH)_4]^{2-}\)
2. \([Zn(OH)_4]^{-}\)
3. \([Zn(OH)_4]^{3-}\)
4. \([Zn(OH)_4]\)

$1.$ ${\mathrm{K}}_2\left[\mathrm{Cr}{\left({\mathrm{C}}_2{\mathrm{O}}_4\right)}_3\right]$
$2.$ ${\mathrm{K}}_3\left[\mathrm{Cr}{\left({\mathrm{C}}_2{\mathrm{O}}_4\right)}_3\right]$
$3.$ ${\mathrm{K}}_3\left[\mathrm{Cr}{\left({\mathrm{C}}_2{\mathrm{O}}_4\right)}_2\right]$
$4.$ ${\mathrm{K}}_2\left[\mathrm{Cr}{\left({\mathrm{C}}_2{\mathrm{O}}_4\right)}_4\right]$