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2 years ago

What's ligand and how are they classified

Chemistry

1 answer:

KATRIN_1 [288]2 years ago

8 0

Explanation:

Ligands: In co-ordination chemistry ligands are ion, molecule or any species which donates electron pair to central metal atom.

Depending on the type of interaction Ligands are of three types.

Sigma donor only
sigma as well as pi donor
pi acceptor ligand

let's understand each type of Ligands individually & in more detail.

1 - Sigma donor only: This is a unidirectional interaction, in which filled ligand overlaps (head to head) with central metal atom/ion & donates pair of electron in the LUMO of metal.

generally all the molecules of 2nd period without pi bond comes in this category, below are few example of sigma donor ligands,

$\small \sf NH_3, H_2O, CH_3^-, H^-, R-OH, R-NH_3, etc$

2- Pi donor: This in also a unidirectional interaction between ligand & central metal atom but the along with head to head overlap, side overlapping takes place.

generally protonated neutral molecules who have more than one pair to donate show such interaction, for e.g.

NH3 have two lone pair to donate but the energy level of both the lone pairs are different hence when it is neutral it only donates one pair of electron. but when NH3 is protonated to NH2- it have two electron pairs (negative charge+ lone pair) to donate & both the pairs have same energy level. example of such ligands are below,

$\sf \small NH_2^-, OH^-, R-O^-, R-NH^-, F^-, Cl^-, Br^- SH^- etc$

3- Pi acceptor ligand: This is a bidirectional interaction between ligand & central metal atom/ion, the filled orbital of ligand undergoes head to head to overlap with vacant orbital of central metal atom, & filled D orbital of central metal donates their pair to vacant LUMO of ligand.

depending on the LUMO pi acceptor ligands are further classified into two categories.

$d\pi - \sigma* \small \sf When \: lumo \: is \: \sigma*\\ d\pi - \pi* \small \: \sf When \: lumo \: is \: \pi*$

The dπ-σ* is seen in molecules of 3rd period onwards without pi bond for e.g.

PH3, PR3, AsR3 & SR2 etc

The dπ-π* is seen in molecules of 2nd or3rd period with pi bond for e.g.

CO C N- SC N^- etc

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Svetach [21]

Answer:

The solution 0.010 M has a higher percent ionization of the acid.

Explanation:

The percent ionization can be found using the following equation:

$\% I = \frac{[H_{3}O^{+}]}{[CH_{3}COOH]} \times 100$

Since we know the acid concentration in the two cases, we need to find [H₃O⁺].

By using the dissociation of acetic acid in the water we can calculate the concentration of H₃O⁺ in the two cases:

1. Case 1 (0.1 M):

CH₃COOH(aq) + H₂O(l) ⇄ CH₃COO⁻(aq) + H₃O⁺(aq) (1)

0.1 - x x x

$Ka = \frac{[CH_{3}COO^{-}][H_{3}O^{+}]}{[CH_{3}COOH]}$ (2)

Where:

Ka: is the dissociation constant of acetic acid = 1.7x10⁻⁵.

$1.7 \cdot 10^{-5} = \frac{x^{2}}{0.1 - x}$

$1.7 \cdot 10^{-5}*(0.1 - x) - x^{2} = 0$

By solving the above equation for x we have:

x = 1.29x10⁻³ M = [CH₃COO⁻] = [H₃O⁺]

Hence, the percent ionization is:

$\% I = \frac{1.29 \cdot 10^{-3} M}{0.1 M} \times 100 = 1.29 \%$

2. Case 2 (0.01 M):

The dissociation constant from reaction (1) is:

$Ka = \frac{[CH_{3}COO^{-}][H_{3}O^{+}]}{[CH_{3}COOH]}$

With [CH₃COOH] = 0.01 M

$1.7 \cdot 10^{-5} = \frac{x^{2}}{0.01 - x}$

$1.7 \cdot 10^{-5}*(0.01 - x) - x^{2} = 0$

By solving the above equation for x:

x = 4.04x10⁻⁴ M = [CH₃COO⁻] = [H₃O⁺]

Then, the percent ionization for this case is:

$\% I = \frac{4.04 \cdot 10^{-4} M}{0.01 M} \times 100 = 4.04 \%$

As we can see, the solution 0.010 M has a higher percent ionization of the acetic acid.

Therefore, the solution 0.010 M has a higher percent ionization of the acid.

I hope it helps you!

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3 years ago

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Answer:

mercury, magnesium, water and gasoline.

Explanation:

Density:

Density is equal to the mass of substance divided by its volume.

Units:

SI unit of density is Kg/m3.

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iVinArrow [24]

Answer:

<h3>The answer is 6 moles</h3>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

$n = \frac{N}{L} \\$

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

$n = \frac{3.612 \times {10}^{24} }{6.02 \times {10}^{23} } \\$

We have the final answer as

<h3>6 moles</h3>

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3 years ago

What's ligand and how are they classified​

What's ligand and how are they classified