Coordination Compounds

To determine the atomic number of the metal based on the provided spin-only magnetic moment value, we use the formula for calculating the spin-only magnetic moment: $\mu = \sqrt{n(n+2)} \mu_{B}$ where $\mu$ is the magnetic moment in Bohr magnetons ($\mu_{B}$), and $n$ is the number of unpaired electrons.

Given that the spin-only magnetic moment value is $3.86 \, \mathrm{BM}$, we can set this value equal to the formula to solve for $n$: $3.86 = \sqrt{n(n+2)}$ Squaring both sides gives: $14.8996 = n(n+2)$ Since $n$ must be an integer and this equation doesn't solve neatly for an integer $n$, we look for a value of $n$ that would give a product close to $14.8996$ when plugged into $\sqrt{n(n+2)}$.

Practically, we can test the square root values near $3.86$ for different $n$ to see which one gives a close match.

Considering the usual spin-only magnetic moments for common numbers of unpaired electrons: $n = 1$, $\mu = \sqrt{3} \approx 1.73 \mu_B$ $n = 2$, $\mu = \sqrt{8} \approx 2.83 \mu_B$ $n = 3$, $\mu = \sqrt{15} \approx 3.87 \mu_B$ $n = 4$, $\mu = \sqrt{24} \approx 4.90 \mu_B$ The value $n = 3$ gives a magnetic moment of approximately $3.87 \mu_B$, which is very close to the given value, $3.86 \mu_B$.

Therefore, the metal has 3 unpaired electrons in its d-orbital configuration.

In the +2 oxidation state, metals lose electrons from the s-orbital before the d-orbital.

Given that the element is a first row transition metal, starting with scandium (Sc) at atomic number 21 which has an electronic configuration of $[Ar] 3d^1 4s^2$ in its neutral state, we can deduce the configurations: $Z = 22$ for Titanium (Ti), which has a neutral configuration of $[Ar] 3d^2 4s^2$.

In the +2 oxidation state, it would have an electronic configuration of $[Ar] 3d^2$, resulting in 2 unpaired electrons. $Z = 23$ for Vanadium (V), with a neutral configuration of $[Ar] 3d^3 4s^2$.

In the +2 state, its configuration would be $[Ar] 3d^3$, presenting 3 unpaired electrons, which matches our calculation. $Z = 26$ for Iron (Fe), indicating a neutral configuration of $[Ar] 3d^6 4s^2$.

In the +2 state, $[Ar] 3d^6$, which would have 4 unpaired electrons, not matching the calculation. $Z = 25$ for Manganese (Mn), with a neutral configuration of $[Ar] 3d^5 4s^2$.

In the +2 state, $[Ar] 3d^5$, indicating 5 unpaired electrons, which also does not match our calculation.

So, the atomic number of the metal with a +2 oxidation state and a spin-only magnetic moment value of $3.86 \mathrm{~BM}$ (corresponding to 3 unpaired electrons) is 23, which identifies the metal as Vanadium (V).

Therefore, Option B (23) is the correct answer.

To determine the number of complexes with an even number of electrons in the

orbitals, we first need to determine the electronic configuration of the metal ions in each complex and then find the distribution of electrons among the

and

orbitals. Let's consider each complex one by one: 1.

The oxidation state of Fe is +2. The electronic configuration of

is

. In an octahedral field, the splitting pattern for the 3d orbitals is such that:

(lower energy) and

(higher energy). So, the electronic configuration in octahedral field will be:

. Therefore, there are 4 electrons in the

orbitals (even). 2.

The oxidation state of Co is +2. The electronic configuration of

is

. In an octahedral field, the splitting pattern for the 3d orbitals will be:

. Therefore, there are 5 electrons in the

orbitals (odd). 3.

The oxidation state of Co is +3. The electronic configuration of

is

. In an octahedral field, the splitting pattern for the 3d orbitals will be:

. Therefore, there are 4 electrons in the

orbitals (even). 4.

The oxidation state of Cu is +2. The electronic configuration of

is

. In an octahedral field, the splitting pattern for the 3d orbitals will be:

. Therefore, there are 6 electrons in the

orbitals (even). 5.

The oxidation state of Cr is +2. The electronic configuration of

is

. In an octahedral field, the splitting pattern for the 3d orbitals will be:

. Therefore, there are 3 electrons in the

orbitals (odd). So, the complexes with an even number of electrons in the

orbitals are: 1.

(4 electrons)

(4 electrons)

(6 electrons) Hence, there are 3 complexes with an even number of electrons in the

orbitals. Therefore, the correct answer is option A.

To solve this problem, we need to determine the oxidation state of cobalt (

) and the number of ammonia molecules (

) in the complex

, given that it produces 2 moles of

upon reaction with excess

. First, let's write the reaction between the complex and

. The precipitate formation indicates that some chloride ions are free (not coordinated to the metal). Since 2 moles of

are formed, it indicates that there are 2 chloride ions that are free to react with

.

Therefore, we can conclude that in the complex, 1 chloride ion is coordinated to the cobalt, and 2 chloride ions are free.

This can be represented as:

Now, let's determine the oxidation state of cobalt (Co).

The charge on the entire complex should be zero, so we can write the charge balance equation as follows: The sum of the charges is:

Solving this, we get:

So, the oxidation state of Co is 3. Now, we need to find the value of

.

Since the coordination number of Co in an octahedral complex is typically 6 and we have 1 chloride ion coordinated to Co, the number of ammonia molecules coordinated to Co would be:

Finally, we calculate

:

So, the value of "

" is 8. Therefore, the correct option is: Option D: 8

Answer: Option D - Statement I is correct but Statement II is incorrect. Explanation: Statement I:

(trimethylamine) and

(trimethylphosphine) can indeed act as ligands to form transition metal complexes.

Ligands are molecules or ions that can donate a pair of electrons to a metal atom to form a coordinate bond.

Both nitrogen and phosphorus have lone pairs of electrons that can be donated to transition metals, making

and

suitable ligands.

Therefore, Statement I is correct.

Statement II: While N (Nitrogen) and P (Phosphorus) are both from group 15 of the periodic table and share some similar properties, the nature of their bonding with transition metals is not always the same.

Nitrogen compounds typically form stronger bonds with metals compared to phosphorus compounds.

This difference is due to several factors, including the difference in atomic sizes, electronegativity, and the extent of orbital overlap.

Nitrogen, being smaller and more electronegative, can form more effective overlap and stronger bonds with metals compared to phosphorus.

Therefore, Statement II is incorrect.

Given this analysis, the most appropriate answer is Option D: Statement I is correct but Statement II is incorrect.

The correct answer is Option C:

\mathrm{F}^{-}

\mathrm{Br}^{-}

\mathrm{CN}^{-}

\mathrm{NH}_3

\mathrm{CN}^{-}

\mathrm{NH}_3

\mathrm{CN}^{-}

\mathrm{Br}^{-}

\mathrm{Br}^{-}

So, the complex with minimum number of unpaired

is option (2)

.

Identifying the electronic configuration and geometry : The condition "no electrons in the $t_2$ orbital" typically applies to a tetrahedral crystal field splitting pattern.

In a tetrahedral field: The $d$-orbitals split into two sets: $e$ (lower energy, 2 orbitals) and $t_2$ (higher energy, 3 orbitals).

Electrons fill the lower-energy $e$ set before occupying the $t_2$ set.

Thus, to have no electrons in $t_2$, either: The metal has a $d^0$ configuration (no $d$-electrons at all), or The $d$-electron count is so low that all electrons can fit into the $e$ orbitals without needing to occupy $t_2$.

Analyzing each complex : (i) $\mathrm{TiCl}_4$ : Ti in TiCl₄ is in the +4 oxidation state (since TiCl₄ is neutral and each Cl is -1).

Ti (Z = 22) neutral : $3d^2 4s^2$.

As Ti⁴⁺, it loses all 4 valence electrons (2 from 4s and 2 from 3d), resulting in $d^0$.

With $d^0$, there are no electrons to occupy $t_2$ orbitals.

No electrons in $t_2$. (ii) $[\mathrm{MnO}_4]^{-}$ (permanganate) : Mn in permanganate ($\mathrm{MnO_4}^-$) is in the +7 oxidation state.

Mn (Z = 25) neutral is $3d^5 4s^2$.

Mn⁷⁺ means removing all 7 valence electrons, leaving $d^0$.

With $d^0$, no electrons in $t_2$.

No electrons in $t_2$. (iii) $[\mathrm{FeO}_4]^{2-}$ (ferrate) : Fe in $\mathrm{FeO_4}^{2-}$: Oxygen contributes -8 total.

The ion is -2.

Thus, Fe + (-8) = -2 → Fe = +6 oxidation state.

Fe (Z = 26) neutral is $3d^6 4s^2$.

Fe⁶⁺ means removing 6 electrons from the valence shell.

After losing 2 from 4s and 4 from 3d, we get $d^2$.

In a tetrahedral field, $d^2$ fills the lower $e$ orbitals (2 electrons into e orbitals).

No need to occupy $t_2$ orbitals since both electrons fit into e.

No electrons in $t_2$. (iv) $[\mathrm{FeCl}_4]^{-}$ : Fe in $\mathrm{FeCl_4}^-$: Cl total charge = -4, complex = -1, so Fe = +3.

Fe(III) is $d^5$.

In a tetrahedral field, with 5 $d$-electrons, after filling the 2 $e$ orbitals, we have 3 more electrons that must go into $t_2$.

Has electrons in $t_2$. (v) $[\mathrm{CoCl}_4]^{2-}$: Co in $\mathrm{CoCl_4}^{2-}$ : Cl total = -4, complex = -2, so Co = +2.

Co(II) is $d^7$.

For $d^7$, even after filling the $e$ set (2 orbitals), we have 5 more electrons left, which must occupy the $t_2$ orbitals.

Has electrons in $t_2$.

Counting the complexes with no electrons in $t_2$ : TiCl₄: No electrons in $t_2$.

[MnO₄]⁻: No electrons in $t_2$.

[FeO₄]²⁻: No electrons in $t_2$.

[FeCl₄]⁻: Has electrons in $t_2$.

[CoCl₄]²⁻: Has electrons in $t_2$.

Total with no electrons in $t_2$ = 3.

Answer : 3

To determine the number of geometrical isomers for the given complex

, where

is a metal and

,

,

, and

are unidentate ligands, we need to consider the geometry implied by the $\mathrm{sp}^3$ hybridization of the metal.

The $\mathrm{sp}^3$ hybridization suggests a tetrahedral geometry for the metal complex.

In a tetrahedral complex, geometrical isomerism does not arise.

This is because geometrical isomerism (also known as cis-trans isomerism) typically occurs in square planar or octahedral complexes where ligands can be positioned differently around the central metal atom to give distinct isomers.

In a tetrahedral complex, every position is equivalent due to the symmetric arrangement of the ligands around the metal center, making it impossible to distinguish between different sides of the molecule as you would with a square planar or octahedral complex.

Given that the complex follows tetrahedral geometry because of $\mathrm{sp}^3$ hybridization and that tetrahedral complexes do not exhibit geometrical isomerism, the number of geometrical isomers exhibited by the complex is: Option A: 0 This is because in a tetrahedral arrangement of ligands around a metal center, no geometrical isomers can be formed due to the lack of cis or trans arrangements possible within such a symmetric geometry.

Homoleptic complexes are complexes with identical ligands attached to the central metal atom. (A)

- Homdeptic Central metal : Fe IN this complex, Fe is in +6 oxidation state. {oxidation state :

Configuration of

Configuration of

Iron has even number of d electrons. So, it is not correct. (B)

- Homoleptic Central metal : Fe In this complex, the oxidation state of Fe is +3 Oxidation state :

Configuration of

Configuration of

Iron has odd number of d electrons. So, it is correct. (C)

- Not homoleptic This complex has different type of ligands.

- Homoleptic Central metal - Co In this complex, the oxidation state of Co is +2 Oxidation state:

Configuration of

Configuration of

Cobalt has odd number of d electrons. So, it is correct. (E)

- Not homoleptic, It has different type of ligands.

Complexes B and D are homoleptic complexes with odd number of d electrons in the central metal.

So, correct answer is (B) and (D) only.

Spin-only magnetic moment formula is

, where n is the number of unpaired electrons in the central metal

Also written as

Oxidation state of

Configuration of

Configuration of

Since OH$^-$ is a weak ligand, no pairing of electrons occurs. Unpaired electrons,

So, spin-only magnetic moment

BM

Also written as

Oxidation state of

Configuration of

Configuration of

Since OH$^-$ is a weak field ligand, no pairing of electrons occurs. Unpaired electrons = 4 Spin-only magnetic moment

BM Spin-only magnetic moments of

and

respectively are 5.92 and 4.90 B.M. Correct answer is option (3) 5.92 and 4.90 B.M.