Chemistry

Explanation:

The atomic number of zinc is 30. The electronic configuration of Zn is \([Ar]\ 3d^{10}\ 4s^2\).  It is a slightly brittle metal at room temperature and has a shiny-greyish appearance when oxidation is removed. It is the first element in group 12 (IIB) of the periodic table.

It is not considered a transition element because it does not have incompletely filled d-orbitals in its ground state. Transition metals form stable ions with incompletely filled d orbitals. It is present in the d-block, but it is not a transition element. 

Both (A) and (R) are true, and (R) is the correct explanation of (A). Thus, option 1 is the correct choice. 
 

Explanation:

(i) Nitriles can be hydrolyzed to carboxylic acids in acidic aqueous solutions. In the first stage, it forms the corresponding amide group, and in the second stage, it forms the corresponding carboxylic acid from the amide.

(ii) Carboxylic acids having α-hydrogen are halogenated at α-position on treatment with Br₂ in the presence of red phosphorous to give \(\alpha\)-halocarboxylic acids. This is the Hell-Volhard-Zelinsky reaction.

So, the final product will be a \(\alpha\)-bromocarboxylic acid. The complete reaction is as follows: 

So, option 2 is the correct choice. 

Explanation:

In case of alkyl halide, the leaving group order is: Iodide > Bromide > Chloride

Iodide is the best leaving group than other halogen atoms due to its larger size. Due to larger size, charge density decreases and it becomes stable. 

So, the order of S_N2 reaction is:  RI > RBr > RCl.  Because as the atoms get bigger from Cl to I , bonds get longer, so weaker and react faster. Length and strength dominate, not polarity of the carbon-halogen bond, which would give the opposite order.
 

In metal carbonyl, synergic bonding takes place. The metal-carbon bond in metal carbonyls possess both σ and π character. The M–C σ bond is formed by the donation of lone pair of electrons on the carbonyl carbon into a vacant orbital of the metal.
The M–C π bond is formed by the donation of a pair of electrons from a filled d orbital of metal into the vacant antibonding π* orbital of carbon monoxide.   

The diagram is as follows:

Thus, the assertion is correct but the reason is false. 

Explanation:
The reason (R) provides the molecular weight of oxygen. The calculation for the volume of oxygen evolved in (A) depends on Faraday’s law and the volume occupied by one mole of gas at STP (22.4 L/mol), which is a consequence of Avogadro’s Law and the Ideal Gas Law. While the molecular weight (32) is consistent with \(O_2​\) occupying 22.4 L/mol at STP, it is not the fundamental chemical or physical principle required to calculate the 11.2 L value based on the 2 F charge. The calculation primarily relies on the stoichiometry of the half-reaction and Faraday’s Laws.

Both (A) and (R) are True, but (R) is not the correct explanation of (A)

Explanation:
The formula of work done is as follows:

\(w=-2.303nRTlog\frac{P_{i}}{P_{f}} \)

The given values are as follows:

n = 1 mole
P_i = 5
P_f = 1
R = 8.314
T=300 K

Calculate the value of work done as follows:

\(w=-2.303\times 1\times 8.314\times 300log\frac{5}{1} \)
\( =-4014.6 \ J \)
 
Hence, the correct option is the fourth option.

Grignard reagent with aldehydes reacts together to give organo-metallic compounds which on hydrolysis results in the formation of alcohol.
In the first step, formaldehyde was reacted with methyl magnesium bromide (Grignard reagent) which gives unstable adduct.

In the second step hydrolysis of product formed in step 1 gives us propan-1-ol(propanol) as a final product.

The reaction can be given as follow:

Sucrose is converted to glucose and fructose in the presence of an enzyme, invertase or sucrase. Glucose and fructose undergo fermentation in the presence of another enzyme, zymase. Both these enzymes are present in yeast.

The reactions are as follows:

Hint: The reaction of nitrogen and oxygen depends upon the temperature.
Both ${\mathrm{N}}_2<mspace\; width="1em"></mspace>\mathrm{and}<mspace\; width="1em"></mspace>{\mathrm{O}}_2$ are less reactive at ordinary conditions. The reaction between nitrogen and oxygen requires a high temperature. N₂ bond order is 3 and it is non reactive at room temperature. Thus, it will not reacts with O₂ 

The bond dissociation energy of N₂ and O₂  is as follows:

N₂ = 945 kJ/mol
O₂ = 498 kJ/mol

Hence, both assertion and reason are correct, and the reason is the correct explanation for the assertion.

Hint: The formula of equilibrium constant is, \(K_{equ} = \frac{Z^{2}}{X\times Y}\)

Step 1:

The concentration of X, Y and Z is as follows:

Concentration = \(=\frac{number \ of \ mole}{volume \ of \ solution}\)

Hence, volume of solution is 1 L, concentration is equal to number of moles. So concentration of X, Y and Z is equal to its mole.

               X   +  Y  =  2Z
t = 0        1      1.5     0.5
At eq.     1-x   1.5-x      0.5 + 2x

The equilibrium concentration of Z is 1. Calculate the value of x as follows:

[Z] = 0.5 + 2x

\(\frac{1-0.5}{2}\)

x = 0.25 M

The concentration of X and Y at equilibrium is as follows:

[X] = 1 – x = 0.75, [Y] = 1.5 – x = 1.25

Step 2:

\(K_{equ} = \frac{Z^{2}}{X\times Y}\)

$K_{eq.}<mspace\; width="1em"></mspace>=\frac{1}{{\displaystyle \frac{3}{4}}\times {\displaystyle \frac{5}{4}}}<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>\frac{16}{15}$