SOLUTION

HI cannot be prepared by the action of conc. H₂SO4 on KI because H₂SO₄  is an oxidising agent and HI is strong reducing agent.

HI reduces sulphuric acid to sulphur dioxide and itself is oxidised to iodine.

KI+H₂SO₄→KHSO₄+HI

H₂SO₄+2HI→SO₂+I₂+2H2O

Hence, HI is prepared by heating KI with conc. phosphoric acid.  Same is true for the preparation of HBr.

SOLUTION
Smaller the bond angle. greater is the resultant dipole moment 

The dipole moment of 2 dipoles inclined at an angle θ is given by,

μ =√ (x²+y²+2xycosθ?)

cos90=0

∴   The dipole moment is maximum for θ=90⁰

Solution

Max^m no. of zero values is given by total no. of nodes which is given by ⇒n−1.

n−1 principle quantum number

Solution

I-propyne reacts with methyl magnesium bromide to form prop-1-ynylmagnesium bromide which is compound A. Methane is the byproduct. The prop-1-ynyl anion acts as a nucleophile and attacks carbon dioxide. This is followed by hydrolysis to form but-2-ynoic acid which is compound B.
$$\displaystyle CH_{3}-C\equiv C-H\xrightarrow{CH_{3}MgBr}CH_{4}+ \underset {\left ( A \right )} {CH_{3}-C\equiv C-MgBr}\xrightarrow[\left ( ii \right )H_{2}O/H^{\oplus } ]{\left ( i \right )CO_{2}}\underset {\left ( B \right )} {\displaystyle CH_{3}-C\equiv C-COOH}$$

SOLUTION

Aldhyde behaves as a functional group and numbering starts from terminal functional group.

2−Hydroxy−4−Oxopentanal

SOLUTION

The fourth IE is significantly higher than third IE. Hence the element attains inert configuration on removal of 33 electrons i.e., it belongs to either IIIA  or IIIB

As only such option is of Al (IIIA). 

SOLUTION

The oxidation number of H=+1

Let the oxidation number of Cl be x..

(+1)×1+(+x)×1+(−2)×4=0

⇒x=8−1=+7

∴ Oxidation number of Cl is +7 .

Solution: 

As NH₄Cl is a salt of strong acid and weak base, it's pH will be less than 7 since the salt is acidic in nature.