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S_A_V [24]
2 years ago
8

What's ligand and how are they classified​

Chemistry
1 answer:
KATRIN_1 [288]2 years ago
8 0

Explanation:

<u>Ligands:</u> 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.

  1. Sigma donor only
  2. sigma as well as pi donor
  3. 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 <em>for e.g.</em>

<em>PH3,</em><em> </em><em>PR</em><em>3</em><em>,</em><em> </em><em>AsR</em><em>3</em><em> </em><em>&</em><em> </em><em>SR</em><em>2</em><em> </em><em>etc</em>

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

CO C N- SC N^- etc

<em><u>Thanks for joining brainly community!</u></em>

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Answer:

0.628 M.

Explanation:

In order to solve this problem we need to keep in mind the<em> definition of molarity</em>:

  • Molarity = moles / liters

We are given both the <em>number of moles and the volume of solution</em>, meaning we can now proceed to <u>calculate the molarity</u>:

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3 years ago
How much energy must be supplied to break a single 21Ne nucleus into separated protons and neutrons if the nucleus has a mass of
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Answer:

1)There is 2.68 * 10^-11 J of energy needed

2) The nuclear binding energy for 1 mol of Ne is 1.6 *10^13 J/mol

Explanation:

Step 1: Data given

The nucleus of a21Ne atom has a amass of 20.98846 amu.

Step 2: Calculate number of protons and neutrons

The number of electrons and protons in an 21Ne atom = 10

The number of neutrons = 21 -10 =11

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The mass of the atom = mass of all neutrons + mass of protons

Mass of atom = 11*1.0086649 + 10*1.00727647  = 21.1680786 amu

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The change in mass = Mass of atom - mass of neon

Δmass = 21.1680786 - 20.98846

Δmass = 0.1796186

Step 5: Calculate mass for a single nucleus

The change of mass for a single nucleus is = Δmass / number of avogadro

Δmass of nucleus = 0.1796186 / 6.022*10^23

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Step 6: Calculate energy to break a Ne nucleus

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ΔE= 2.68 * 10^-11 J * 6.022*10^23

ΔE= 1.6 *10^13 J/mol

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3 years ago
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