Chemistry Test - 9

State of hydrogen atom (n) = 1 (due to ground state)

Radius of hydrogen atom (r) = 0.53 A

Atomic number of Li (Z) = 3

Radius of Li²⁺ ion

$=n\frac{n^2}{2}=0.53\times \frac{{\left(1\right)}^2}{3}=0.17$

(A) g orbital will have n = 5, l = 4

(B) It has 18 electrons because

m = -4, -3, -2, -1, 0, +1, +2, +3, +4 i.e., 9 value 9 orbitals therefore 9 × 2 = 18e

(C) Orbital will be of 9 types because m, has 9 value given in (B)

Hence option D is the correct answer.

The energy required to remove an electron from an isolated gaseous atom in its ground state is known as ionisation energy. The closer and more tightly bound an electron is to the nucleus, the more difficult it will be to remove, and the higher its ionization energy will be. Ionization energies increase on moving from left to right across a period (decreasing atomic radius). As we move along a period, the nuclear charge increases and the electrons are added into the same shell. Thereby the effective nuclear charge increases and size decreases. Therefore, the energy required to remove an electron increases.

Ionization energy decreases down a group in the periodic table due to the fact that the outermost electrons are further away from the nucleus. The increase in distance between the positive nucleus and the negative electrons creates a weaker attraction between the two. As a result, less energy is needed to remove an electron from elements located lower on the periodic table.

Ionisation energy depends on:

Atomic radius- Larger the atom lesser the attractive force felt by the valence electrons. Hence lesser energy is required to pull out an electron . Thus ionisation energy is small. For smaller atoms reverse is true. Smaller atoms have higher ionisation energy.

Nuclear charge- If the nuclear charge is more the attractive force on the valence electron is more. The energy required to pull out the valence electron will be high and ionisation energy will be high.

Stability of half filled and completely filled orbitals- The atoms with fully filled and half filled orbitals have greater stability than the others. Therefore they require greater energy for removing an electron.

Here the electronic configuration of atomic number of nitrogen is 15 and its outer electronic configuration is 2s² 2p³. Oxygen has atomic number 16 and its outer electronic configuration is 2s² 2p⁴. 2s² 2p³ is more stable than 2s² 2p⁴ due to half filled p sublevel. Hence nitrogen atom has greater ionisation energy than oxygen.

Hence option D is the correct answer.

$4N{H}_3+5O_2⟶4NO+6H_{2}O$

$1{\mathrm{NH}}_3+1.25{\mathrm{O}}_2⟶1\mathrm{NO}+1.5{\mathrm{H}}_2\mathrm{O}$

Hence 1 mole of NH3 reacts with 125 125 moles of O2 to produce 1 mole of NO and 1.5 moles of H₂O. Hence when one mole of ammonia and one mole of oxygen are made to react to completion, then all the oxygen are made to react to completion, then all the oxygen is consume.

Hence option C is the correct answer.

The number of maxima in 2x orbital is two. (picture attached) also as we know, Total no of nodes are n - 1

The number of spherical or radical nodes is equal to n - l - 1. The number of angular nodes are l.

Hence option D is the correct answer.

Mg = ls² 2s² 2p⁶ 3s²

After removing of 2 electron, the magnesium acquired noble gas configuration hence removing of 3rd electron will require large amount of energy.

Hence option A is the correct answer.

Addition of halogens is stereospecific reaction. Trans-z-butene gives meso product whereas the cis-isomer gives racemic mixture.

The following statements are correct statements

(A) Bromination proceeds through trans-addition in both the reactions. It involces formation of cyclic bromonium ion.

(C) (M and O) are two pairs of diastereomers. They have same configuration at one chiral carbon atom and different configuration at other chiral carbon atom.

The following are incorrect statements

(D) "(M and O) and (N and P) are two pairs of enantiomers". The correct statement is "(M and O) and (N and P) are two pairs of diastereomers".

Hence option D is the correct answer.

∆G₀ = -2.303 RT log K

∆G₀ = -2.303 RT log K

For a reversible reaction at constant temperature and pressure,∆G = ∆G₀ + RT ln Q

At equilibrium, ∆G = 0, Q = K

Therefore 0 = ∆G₀ + RT ln K

Therefore ∆G₀ = - RT ln K

or ∆G₀ = -2.303 RT log K

Hence option A is the correct answer.

Compounds A and B are optically inactive, as one isomer can rapidly interconvert into another isomer (mirror image).

In compound C the interconversion of one isomer into another isomer (mirror image) is not possible. Also, different groups are attached to the central N atom in both of these compounds. So, compound C is optically active.

Hence option C is the correct answer.

Molecular Orbital Theory (MOT):

Basic idea of MOT is that atomic orbitals of individual atoms combine to form molecular orbitals. Electrons in molecule are present in the molecular orbitals which are associated with several nuclei.

The molecular orbital formed by the addition of atomic orbitals is called the bonding molecular orbitals.

The molecular orbital formed by the subtraction of atomic orbital is called anti-bonding molecular orbitals.

The sigma (s) molecular orbitals are symmetrical around the bond-axis while pi (p) molecular orbitals are not symmetrical.Bond order(b.o.) is defined as one half the difference between the number of electrons present in the bonding and theanti-bonding orbitals.So all A,B & C are basic principles in MO theory.

Hence option D is the correct answer.

According to VSEPR theory , in a T – shaped molecule of the type MX₃ , there will be two lone pairs oriented at equatorial positions . eg : ClF . The central atom (M) undergoes sp3d hybridisation

Hence option B is the correct answer.

Given that,  200ml of 1M KOH is added to 200 ml of 1M HCl and themixture is well shaken. 

Rise in temperature = T₁ 

100 ml of each solution is again mixed.

T₂=?

As when quantity is halved, heat released will also be halved.

So, T₁ = T₂

Hence option C is the correct answer.

For 3 bond pairs and 2 lone pairs, hybridization is sp³d.

Hence, trigonal bipyramidal geometry.

The geometry is also called as T-shaped geometry. Example: ClF₃

Hence option A is the correct answer.