= 1.125 atm⁻²

Since value of Q_p is larger than K_p (4.28 × 10⁻⁵ atm⁻²), it indicates net reaction will proceed in backward direction.

The four lobes of  orbital are lying along x and y axes, so, density in XY plane can't be zero. 

The two lobes of  orbital are lying along z axis, and contain a ring of negative charge surrounding the nucleus in xy plane. So, the density in XY plane is non-zero.

2s orbitals have one spherical node, where the electron density is zero.

In 2p_x orbital, both the lobes lie along x axis. Hence,the density in yz plane is zero, thus it is the nodal plane.

Higher the electrostatic attraction between nucleus and valence electron, smaller will be the size of the atom/ion.

Electrostatic attraction is given by Z_eff = Z−S where, Z= the number of protons in the nucleus of an atom or ion (the atomic number) and S= shielding from core electrons.

In case of Al³⁺ and Mg²⁺, they are isoelectronic species and thus have same number of electrons as 1s²,2s²,2p⁶ isoelectric series of atoms and ions with different numbers of protons (and thus different nuclear attraction, gives) the relative ionic sizes of each atom or ion with respect to atomic number.

Atomic number, Z_Al = 13 and ZMg = 12.

Thus, Al³⁺ has lower ionic radii than Mg²⁺ due to higher nuclear charge.

Hence, A is true, R is true and R is the correct explanation for A.

The structure with more delocalization and more charge separation is the most stable.

• Structure I:
  Formal charges: Left = -1, Middle = +1, Right = -1

• Structure II:
  Formal charges: Left = -2, Middle = +1, Right = 0

• Structure III:
  Formal charges: Left = 0, Middle = +1, Right = -2

Conclusion:

• Structure I has more delocalization and charge separation, making it the most stable.
• Structures II and III have equal stability.

Due to delocalisation of π-electrons, (C—Cl) bond is stable and it does not show S_N reactions; Cl directly attached (C=C) bond, i.e. vinyl group.

The correct answer is Option B.

General electronic configuration is given:

(n − 2)f^1-14(n − 1)d⁰¹ns² , where(n=7)

5f^1-146d⁰¹7s²

The seventh period (n=7) is similar to the sixth period with the successive filling up of the 7s, 5f, 6d and 7p orbitals and includes most of the man-made radioactive elements. This period will end at the element with atomic number 118 which would belong to the noble gas family. Filling up of the 5f orbitals after actinium (Z=89) gives the 5f-inner transition series known as the Actinide Series.