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 :
Let's analyze the trends in Group 13 elements (Boron family) for the given properties: B, Al, Ga, In, Tl. Atomic Radius: Generally, atomic radius increases down a group due to the addition of new electron shells.
However, there's an anomaly due to poor shielding by the d-electrons in Ga.
The correct order is B<Al<Ga<In<Tl. Order A: B<Al<Ga<In<Tl (Correct)
Electronegativity: Electronegativity generally decreases down a group due to increasing atomic size and shielding. However, due to the poor shielding by d-electrons in Ga and In, their electronegativities are slightly higher than expected.
The general trend is B>Al>Ga \approx In>Tl. The given order is Al<Ga<In<Tl<B. Order B: Al<Ga<In<Tl<B (Correct) Density: Density generally increases down a group due to increasing atomic mass. However, the volume also plays a role.
The order is B<Al<Ga<In<Tl. The given order is Tl<In<Ga<Al<B. Order C: Tl<In<Ga<Al<B (Incorrect, the density of Al is less than Ga) First Ionisation Energy: Ionisation energy generally decreases down a group due to increasing atomic size and shielding. However, there are irregularities due to the electronic configurations.
The correct order is B>Al<Ga<Tl<In. The given order is In<Al<Ga<Tl<B. Order D: In<Al<Ga<Tl<B (Correct) From the analysis: Order A (Atomic radius) is correct. Order B (Electronegativity) is correct. Order C (Density) is incorrect. Order D (1st Ionisation Energy) is correct.
The correct orders are B and D. Therefore, the correct option is (1).
Let \[ I(x) = \int \frac{dx}{(x-11)^{\frac{11}{13}} (x+15)^{\frac{15}{13}}} \] If \[ I(37) - I(24) = \frac{1}{4} \left( b^{\frac{1}{13}} - c^{\frac{1}{13}} \right) \] where \( b, c \in \mathbb{N} \), then \[ 3(b + c) \] is equal to:
For the thermal decomposition of \( N_2O_5(g) \) at constant volume, the following table can be formed, for the reaction mentioned below: \[ 2 N_2O_5(g) \rightarrow 2 N_2O_4(g) + O_2(g) \] Given: Rate constant for the reaction is \( 4.606 \times 10^{-2} \text{ s}^{-1} \).