Total number of molecules/species from following which will be paramagnetic is \underline\hspace2cm . O_2, O_2^+, O_2^-, NO, NO_2, CO, K_2[NiCl_4],[Co(NH_3)_6] Cl_3, ~K_2[Ni(CN)_4]

Coordination Compounds PYQ — JEE Main 2025 · 28 Jan · Shift 2 · Q22 with Solution

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Correct answer: 6

List each species and check for unpaired electrons

We need to count how many among the following are paramagnetic:

\mathrm{O_2},\ \mathrm{O_2^+},\ \mathrm{O_2^-},\ \mathrm{NO},\ \mathrm{NO_2},\ \mathrm{CO},\ \mathrm{K_2[NiCl_4]},\ [\mathrm{Co(NH_3)_6}]\mathrm{Cl_3},\ \mathrm{K_2[Ni(CN)_4]}

A species is paramagnetic if it has one or more unpaired electrons.

Analyze each species

(i) $\mathrm{O_2}$

By molecular orbital theory, $\mathrm{O_2}$ has 2 unpaired electrons. So, $\mathrm{O_2}$ is paramagnetic.

(ii) $\mathrm{O_2^+}$

Removing one electron from $\mathrm{O_2}$ leaves 1 unpaired electron. So, $\mathrm{O_2^+}$ is paramagnetic.

(iii) $\mathrm{O_2^-}$

Adding one electron to $\mathrm{O_2}$ pairs one of the unpaired electrons, but still 1 unpaired electron remains. So, $\mathrm{O_2^-}$ is paramagnetic.

(iv) $\mathrm{NO}$

Total electrons = $7+8=15$ , an odd-electron molecule. Hence it has 1 unpaired electron. So, $\mathrm{NO}$ is paramagnetic.

(v) $\mathrm{NO_2}$

Total valence electrons = $5 + 2(6) = 17$ , odd-electron species. Hence it has 1 unpaired electron. So, $\mathrm{NO_2}$ is paramagnetic.

(vi) $\mathrm{CO}$

Isoelectronic with $\mathrm{N_2}$ ; all electrons are paired. So, $\mathrm{CO}$ is diamagnetic.

Coordination compounds

(vii) $\mathrm{K_2[NiCl_4]}$

The complex ion is $[\mathrm{NiCl_4}]^{2-}$ .

Let oxidation state of Ni be $x$ :

x + 4(-1) = -2 \Rightarrow x=+2

So Ni is $\mathrm{Ni^{2+}}$ , with configuration:

\mathrm{Ni}: [Ar]3d^8 4s^2 \Rightarrow \mathrm{Ni^{2+}}: [Ar]3d^8

$\mathrm{Cl^-}$ is a weak field ligand, so $[\mathrm{NiCl_4}]^{2-}$ is tetrahedral and high spin. A tetrahedral $d^8$ complex has 2 unpaired electrons.

Therefore, $\mathrm{K_2[NiCl_4]}$ is paramagnetic.

(viii) $[\mathrm{Co(NH_3)_6}]\mathrm{Cl_3}$

The complex cation is $[\mathrm{Co(NH_3)_6}]^{3+}$ .

Oxidation state of Co:

x + 6(0) = +3 \Rightarrow x=+3

So Co is $\mathrm{Co^{3+}}$ :

\mathrm{Co}: [Ar]3d^7 4s^2 \Rightarrow \mathrm{Co^{3+}}: [Ar]3d^6

$\mathrm{NH_3}$ with $\mathrm{Co^{3+}}$ gives a strong enough field octahedral low-spin complex:

t_{2g}^6 e_g^0

All electrons are paired. So, $[\mathrm{Co(NH_3)_6}]\mathrm{Cl_3}$ is diamagnetic.

(ix) $\mathrm{K_2[Ni(CN)_4]}$

The complex ion is $[\mathrm{Ni(CN)_4}]^{2-}$ .

Oxidation state of Ni:

x + 4(-1) = -2 \Rightarrow x=+2

Again Ni is $\mathrm{Ni^{2+}} = 3d^8$ .

$\mathrm{CN^-}$ is a strong field ligand, causing pairing. For $d^8$ Ni(II), this complex is square planar and all electrons are paired. So, $\mathrm{K_2[Ni(CN)_4]}$ is diamagnetic.

Count the paramagnetic species

Paramagnetic species are:

$\mathrm{O_2}$
$\mathrm{O_2^+}$
$\mathrm{O_2^-}$
$\mathrm{NO}$
$\mathrm{NO_2}$
$\mathrm{K_2[NiCl_4]}$

Total number:

6

Final answer

The total number of paramagnetic species is

\boxed{6}

More from Coordination Compounds

View written solutionFree

Correct answer: 6

List each species and check for unpaired electrons

We need to count how many among the following are paramagnetic:

\mathrm{O_2},\ \mathrm{O_2^+},\ \mathrm{O_2^-},\ \mathrm{NO},\ \mathrm{NO_2},\ \mathrm{CO},\ \mathrm{K_2[NiCl_4]},\ [\mathrm{Co(NH_3)_6}]\mathrm{Cl_3},\ \mathrm{K_2[Ni(CN)_4]}

A species is paramagnetic if it has one or more unpaired electrons.

Analyze each species

(i) $\mathrm{O_2}$

By molecular orbital theory, $\mathrm{O_2}$ has 2 unpaired electrons. So, $\mathrm{O_2}$ is paramagnetic.

(ii) $\mathrm{O_2^+}$

Removing one electron from $\mathrm{O_2}$ leaves 1 unpaired electron. So, $\mathrm{O_2^+}$ is paramagnetic.

(iii) $\mathrm{O_2^-}$

Adding one electron to $\mathrm{O_2}$ pairs one of the unpaired electrons, but still 1 unpaired electron remains. So, $\mathrm{O_2^-}$ is paramagnetic.

(iv) $\mathrm{NO}$

Total electrons = $7+8=15$ , an odd-electron molecule. Hence it has 1 unpaired electron. So, $\mathrm{NO}$ is paramagnetic.

(v) $\mathrm{NO_2}$

Total valence electrons = $5 + 2(6) = 17$ , odd-electron species. Hence it has 1 unpaired electron. So, $\mathrm{NO_2}$ is paramagnetic.

(vi) $\mathrm{CO}$

Isoelectronic with $\mathrm{N_2}$ ; all electrons are paired. So, $\mathrm{CO}$ is diamagnetic.

Coordination compounds

(vii) $\mathrm{K_2[NiCl_4]}$

The complex ion is $[\mathrm{NiCl_4}]^{2-}$ .

Let oxidation state of Ni be $x$ :

x + 4(-1) = -2 \Rightarrow x=+2

So Ni is $\mathrm{Ni^{2+}}$ , with configuration:

\mathrm{Ni}: [Ar]3d^8 4s^2 \Rightarrow \mathrm{Ni^{2+}}: [Ar]3d^8

$\mathrm{Cl^-}$ is a weak field ligand, so $[\mathrm{NiCl_4}]^{2-}$ is tetrahedral and high spin. A tetrahedral $d^8$ complex has 2 unpaired electrons.

Therefore, $\mathrm{K_2[NiCl_4]}$ is paramagnetic.

(viii) $[\mathrm{Co(NH_3)_6}]\mathrm{Cl_3}$

The complex cation is $[\mathrm{Co(NH_3)_6}]^{3+}$ .

Oxidation state of Co:

x + 6(0) = +3 \Rightarrow x=+3

So Co is $\mathrm{Co^{3+}}$ :

\mathrm{Co}: [Ar]3d^7 4s^2 \Rightarrow \mathrm{Co^{3+}}: [Ar]3d^6

$\mathrm{NH_3}$ with $\mathrm{Co^{3+}}$ gives a strong enough field octahedral low-spin complex:

t_{2g}^6 e_g^0

All electrons are paired. So, $[\mathrm{Co(NH_3)_6}]\mathrm{Cl_3}$ is diamagnetic.

(ix) $\mathrm{K_2[Ni(CN)_4]}$

The complex ion is $[\mathrm{Ni(CN)_4}]^{2-}$ .

Oxidation state of Ni:

x + 4(-1) = -2 \Rightarrow x=+2

Again Ni is $\mathrm{Ni^{2+}} = 3d^8$ .

$\mathrm{CN^-}$ is a strong field ligand, causing pairing. For $d^8$ Ni(II), this complex is square planar and all electrons are paired. So, $\mathrm{K_2[Ni(CN)_4]}$ is diamagnetic.

Count the paramagnetic species

Paramagnetic species are:

$\mathrm{O_2}$
$\mathrm{O_2^+}$
$\mathrm{O_2^-}$
$\mathrm{NO}$
$\mathrm{NO_2}$
$\mathrm{K_2[NiCl_4]}$

Total number:

6

Final answer

The total number of paramagnetic species is

\boxed{6}

Coordination Compounds question

Coordination Compounds question

(i) $\mathrm{O_2}$

(ii) $\mathrm{O_2^+}$

(iii) $\mathrm{O_2^-}$

(iv) $\mathrm{NO}$

(v) $\mathrm{NO_2}$

(vi) $\mathrm{CO}$

(vii) $\mathrm{K_2[NiCl_4]}$

(viii) $[\mathrm{Co(NH_3)_6}]\mathrm{Cl_3}$

(ix) $\mathrm{K_2[Ni(CN)_4]}$

More from Coordination Compounds

Coordination Compounds question

Coordination Compounds question

(i) $\mathrm{O_2}$

(ii) $\mathrm{O_2^+}$

(iii) $\mathrm{O_2^-}$

(iv) $\mathrm{NO}$

(v) $\mathrm{NO_2}$

(vi) $\mathrm{CO}$

(vii) $\mathrm{K_2[NiCl_4]}$

(viii) $[\mathrm{Co(NH_3)_6}]\mathrm{Cl_3}$

(ix) $\mathrm{K_2[Ni(CN)_4]}$

More from Coordination Compounds

(i) O2\mathrm{O_2}O2​

(ii) O2+\mathrm{O_2^+}O2+​

(iii) O2−\mathrm{O_2^-}O2−​

(iv) NO\mathrm{NO}NO

(v) NO2\mathrm{NO_2}NO2​

(vi) CO\mathrm{CO}CO

(vii) K2[NiCl4]\mathrm{K_2[NiCl_4]}K2​[NiCl4​]

(viii) [Co(NH3)6]Cl3[\mathrm{Co(NH_3)_6}]\mathrm{Cl_3}[Co(NH3​)6​]Cl3​

(ix) K2[Ni(CN)4]\mathrm{K_2[Ni(CN)_4]}K2​[Ni(CN)4​]

More from Coordination Compounds

(i) O2\mathrm{O_2}O2​

(ii) O2+\mathrm{O_2^+}O2+​

(iii) O2−\mathrm{O_2^-}O2−​

(iv) NO\mathrm{NO}NO

(v) NO2\mathrm{NO_2}NO2​

(vi) CO\mathrm{CO}CO

(vii) K2[NiCl4]\mathrm{K_2[NiCl_4]}K2​[NiCl4​]

(viii) [Co(NH3)6]Cl3[\mathrm{Co(NH_3)_6}]\mathrm{Cl_3}[Co(NH3​)6​]Cl3​

(ix) K2[Ni(CN)4]\mathrm{K_2[Ni(CN)_4]}K2​[Ni(CN)4​]

More from Coordination Compounds

(i) $\mathrm{O_2}$

(ii) $\mathrm{O_2^+}$

(iii) $\mathrm{O_2^-}$

(iv) $\mathrm{NO}$

(v) $\mathrm{NO_2}$

(vi) $\mathrm{CO}$

(vii) $\mathrm{K_2[NiCl_4]}$

(viii) $[\mathrm{Co(NH_3)_6}]\mathrm{Cl_3}$

(ix) $\mathrm{K_2[Ni(CN)_4]}$

(i) $\mathrm{O_2}$

(ii) $\mathrm{O_2^+}$

(iii) $\mathrm{O_2^-}$

(iv) $\mathrm{NO}$

(v) $\mathrm{NO_2}$

(vi) $\mathrm{CO}$

(vii) $\mathrm{K_2[NiCl_4]}$

(viii) $[\mathrm{Co(NH_3)_6}]\mathrm{Cl_3}$

(ix) $\mathrm{K_2[Ni(CN)_4]}$