
The correct answer is (A) : 3s2, 3p4
Element E is Selenium
The element which is just above ‘E’ in periodic table is sulphur, its electronic configuration is 1s2, 2s2, 2p6, 3s2, 3p4.
\(E⇒[Ar]^3d^10^4s^24p^4\)
Element above \(E⇒[Ne]3s^23p^4\)
Given below are two statements. 
In the light of the above statements, choose the correct answer from the options given below:
Given below are the pairs of group 13 elements showing their relation in terms of atomic radius. $(\mathrm{B}<\mathrm{Al}),(\mathrm{Al}<\mathrm{Ga}),(\mathrm{Ga}<\mathrm{In})$ and $(\mathrm{In}<\mathrm{Tl})$ Identify the elements present in the incorrect pair and in that pair find out the element (X) that has higher ionic radius $\left(\mathrm{M}^{3+}\right)$ than the other one. The atomic number of the element (X) is
Given below are two statements: 
Statement I: Nitrogen forms oxides with +1 to +5 oxidation states due to the formation of $\mathrm{p} \pi-\mathrm{p} \pi$ bond with oxygen.
Statement II: Nitrogen does not form halides with +5 oxidation state due to the absence of d-orbital in it. 
In the light of the above statements, choose the correct answer from the options given below:
Let $ f(x) = \begin{cases} (1+ax)^{1/x} & , x<0 \\1+b & , x = 0 \\\frac{(x+4)^{1/2} - 2}{(x+c)^{1/3} - 2} & , x>0 \end{cases} $ be continuous at x = 0. Then $ e^a bc $ is equal to