Explain to me how to answer no.2 questions

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First Ionization energy IE 1 for element X = 801

Here X is told to be in the third period.

So principal quantum number n = 3 for X.

For 1st ionization energy the expression is

IE1 = 13.6 x Z ^2 / n^2

Where Z =atomic number.

Thus Z =( n^2 x IE 1 / 13.6)^(1/2)

Z = ( 3^2 x 801 / 13.6 )^ (1/2)

= 23

Number of electrons = Z = 23

Nearest noble gas = Argon

Argon atomic number = 18

Number of extra electrons = 23 – 18 = 5

a)Electronic Configuration= [Ar] 3d34s2

We know that more the value of atomic radii, lower the force of attraction on the electrons by the nucleus and thus lower the first ionization energy.

So more the first ionization energy, less is the atomic radius.

X has more IE1 than Y.

b)So the atomic radius of X is lesser than that of Y.

c)After the first ionization, the atom is no longer electrically neutral. There is an extra proton in the atom. Due to this the remaining electrons are more strongly pulled inside than before ionization. Hence after ionization, the radii of Y decreases.

given the reactants, h2so4 and koh, we are asked to determine the product(s) of the reaction. since h2so4 is an acid and koh is a base. this is a neutralization reaction, thus, the products are salt and water. 

so, h+ from h2so4 will react with oh- from koh forming h2o. k+ will also react with so4 2- forming k2so4. therefore, the balanced reaction will be:

h2so4 + 2koh > k2so4 + 2h2o

explanation:

the positions which points up in upward carbon and down in downward carbon are axial

and vice versa holds true for equatorial positions

so the correct answer with labelled diagram will be