DSSSB PGT Chemistry Mock Test - 9 - DSSSB TGT/PGT/PRT MCQ

V occurs at: 11 ;21 ;24 ;41 ;44 ;
A occurs at: 02 ;03 ;14 ;20 ;31 ;
R occurs at: 65 ;66 ;75 ;89 ;99 ;
Y occurs at: 57 ;58 ;67 ;76 ;97 ;

The statement in option D is incorrect because the passage explicitly states that the embryo is indeed transplanted into a surrogate mother. This is a crucial step in the cloning process.

Since the middle term is not distributed in both the premises, no conclusion can be framed. Secondly, if both the premises begin with "Some" no conclusion can be reached.

To determine the number of trailing zeros in the product of numbers from 1 to 100, we need to count the number of times 10 is a factor in that product. Each 10 is the result of multiplying 2 and 5. Since there are typically more factors of 2 than 5, the number of trailing zeros is determined by the number of times 5 is a factor.

We calculate this by dividing 100 by increasing powers of 5 and taking the integer part of each division:

• Number of multiples of 5: 100 ÷ 5 = 20
• Number of multiples of 25 (which contribute an extra 5): 100 ÷ 25 = 4

Adding these together gives a total of 24 factors of 5, which means there are 24 trailing zeros in the product.

In 1 mole of M_xO,
mole of M²⁺ ion = y
∴ mole of M³⁺ ion = x - y.
Applying charge balance, 2y + 3(x-y)-2 = 0
⇒ y = 3x - 2

Total charge of nickel.
(0.96 x 2) + (0.04 x 3) = 2.04.

Formula of solid = NiO_1.02 = Ni_0.98O.

(c)

Nearest neighbours are along the diagonal at the half distance apart, Diagonal AB = √3a

∴ Distance of the nearest neighbour

Next nearest neighbours are along the cell edge = 520 pm

K-atom at the centre is surrounded by eight K-atoms as nearest neighbours.

In bcc, coordination number is six as next nearest neighbours

The aufbau principle explains how electrons fill low energy orbitals (closer to the nucleus) before they fill higher energy ones. Where there is a choice between orbitals of equal energy, they fill the orbitals singly as far as possible(Hunds rule). The diagram summarizes the energies of the orbitals up to the 4p level.Therefore 3d orbitals have higher energy than the other orbitals in the options.

So answer is: (2) n = 3, l = 2, m = 1, s = +1⁄2

(i) CaOCI₂ has



Oxidation number = + 6
Oxidation number = - 2
(ii) → (p,u)


(iii) → (q.v)

(8) Barium phosphate is Ba₃( P0₄)₂.

Born Haber cycle involves

(a) In SbF₅ (expansion of octet from 5 to 10 and further to 12).
(b) NH₃, BF₃ both have complete octet. NH₃ is electron rich, donates lone pair to BF₃.
(c) AICI₃ is electron deficient, accepts lone pair from Cl^-.
(d) In SF₄, expansion of octet from 8 to 10 and further to 12.

Due to unpaired electron paramagnetic, N-atom in NO₂ is electron deficie nt thus, to complete octet, dimer is formed.

In N₂O₄ formation, each N-atom gets charge. Dipole-dipole repulsion increases (N— N) bond length which is larger than (N— N)covalent bond.

Thus, (a), (b), (c) and (d) are correct. 

The Phosphorus Pentafluoride (PF₅) is a nonpolar molecule because the configuration of PF₅ has trigonal bipyramidal, and PF₅ consists of a central phosphorus atom, surrounded by five fluorine atoms. So the three planar fluorine’s electronegativity cancels out each other, which resulting PF₅ is a nonpolar molecule.

similarly, in PCl₅ case chlorine's electronegativity cancels out each other.

 In case of PCl_3​F₂​, the molecule is non-polar, even though the fluorine atoms are slightly more electro-negative than the chlorine. This could be explain in a couple ways. The more intuitive approach is to consider how the presence of two fluorine atoms would create symmetry. The less intuitive, but more accurate, process would be to realize that the structure of the molecule is divided in to two non-polar structure(trigonal planer and linear). With other molecules, like PF₃​Cl₂​ it's not possible to divide it in to non-polar shapes. The repulsion between the fluorine atom is greater than that between the chlorine. Thus PF_3​Cl_2​ is polar.

Formal charge (F)

where, v = valence electrons = 6
s = shared electrons = 4
u = unshared electron = 4

Number of chiral centers = 2

All four substituent are different so it is chiral compound.

The correct answer is option C
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In halogen-exchange (Finkelstein reaction), reactivity is determined by solubility of Na X formed in acetone. Lower the solubility of Na X in acetone, easier the reaction. NaCI is less soluble than NaBr in acetone.

The reaction involves E2 elimination m echanism in w hich base approaches to β-H from side anti to leaving halide in the rate determining step. Hence, if tertiary butyl group is cis to tosylate, it would not give steric hindrance to the base approaching from opposite side to β-H

Also X, has CH₃—CH(OH)—, gives iodoform test and it is chiral.

It is an α, β-unsaturated ketone which can be formed in an aldol condensation followed by dehydration.