The question involves two statements about the properties of elements in Group 13 and Group 14 of the periodic table. Let's analyze each statement individually to determine their correctness and to substantiate the given answer.
Based on our analysis, we conclude: Statement I is correct but Statement II is incorrect. Hence, the correct option is:
Statement I is correct but Statement II is incorrect
Statement I:
The first ionisation enthalpy of group 14 elements is indeed higher than that of group 13 elements, as the number of valence electrons increases from group 13 to 14, leading to a higher ionisation energy.
Thus, Statement I is correct
Statement II: This statement is incorrect.
The melting points and boiling points of group 13 elements are generally lower than those of group 14 elements due to the stronger metallic bonding in group 14 elements.
Thus, Statement II is incorrect.
Therefore, the correct answer is (1).
Given below are two statements:
Statement I: $ H_2Se $ is more acidic than $ H_2Te $
Statement II: $ H_2Se $ has higher bond enthalpy for dissociation than $ H_2Te $
In the light of the above statements, choose the correct answer from the options given below.
The correct orders among the following are:
[A.] Atomic radius : \(B<Al<Ga<In<Tl\)
[B.] Electronegativity : \(Al<Ga<In<Tl<B\)
[C.] Density : \(Tl<In<Ga<Al<B\)
[D.] 1st Ionisation Energy :
In\(<Al<Ga<Tl<B\)
Choose the correct answer from the options given below :


For the circuit shown above, the equivalent gate is:
Let \( f : \mathbb{R} \to \mathbb{R} \) be a twice differentiable function such that \[ (\sin x \cos y)(f(2x + 2y) - f(2x - 2y)) = (\cos x \sin y)(f(2x + 2y) + f(2x - 2y)), \] for all \( x, y \in \mathbb{R}. \)
If \( f'(0) = \frac{1}{2} \), then the value of \( 24f''\left( \frac{5\pi}{3} \right) \) is: