Chemistry Test - 23

On reaction of haloalkane with methoxide ion, alkene is formed.

Mechanism

\(\stackrel{\ominus}{\mathrm{OCH}_3}\) will act as a base due to its small size and high electron density and therefore, abstracts proton to form double bond which is in conjugation with aromatic ring.

The \(\stackrel{\ominus}{\mathrm{OCH}_3}\) when acts as nucleophile undergoes nucleophilic substitution and replaces \(\mathrm{Br}^{\ominus}\) to form ether, which is a minor product.

So, option (b) is correct.

\(\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+} \Rightarrow \mathrm{Cr}^{2+} \Rightarrow 3 \mathrm{~d}^4 \Rightarrow \mathrm{t}_{2 \mathrm{~g}}^{1,1,1}, \mathrm{e}_{\mathrm{g}}^{1,0}\)

No. of unpaired e : 4

\(\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+} \Rightarrow \mathrm{Fe}^{2+} \Rightarrow 3 \mathrm{~d}^6 \Rightarrow \mathrm{t}_{2 \mathrm{~g}}{ }^{2,1,1}, \mathrm{e}_{\mathrm{g}}^{1,1}\)

No. of unpaired e : 4

\(\left[\mathrm{Fe}\left(\mathrm{NH}_3\right)_6\right]^{2+} \Rightarrow \mathrm{Fe}^{2+} \Rightarrow 3 \mathrm{~d}^6 \Rightarrow \mathrm{t}_{2 \mathrm{~g}}{ }^{2,1,1}=\mathrm{e}_{\mathrm{g}}^{1,1}\)

No. of unpaired e : 4

\(\left[\mathrm{Co}(\mathrm{OH})_4\right]^{2-} \Rightarrow \mathrm{Co}^{2+} \Rightarrow 3 \mathrm{~d}^{\top} \Rightarrow \mathrm{e}^{2,2}, \mathrm{t}_2{ }^{1,1,1}\)

No. of unpaired e: 3

\(\left[\mathrm{CoCl}_4\right]^{2-} \Rightarrow \mathrm{Co}^{2+} \Rightarrow 3 \mathrm{~d}^7 \Rightarrow \mathrm{e}^{2,2}, \mathrm{t}_2^{1,1,1}\)

No. of unpaired \(\mathrm{e}: 3\)

\(\left[\mathrm{Mn}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+} \Rightarrow \mathrm{Mn}^{2+} \Rightarrow 3 \mathrm{~d}^5 \Rightarrow \mathrm{t}_{2 \mathrm{~g}}{ }^{1,1,1}=\mathrm{e}_{\mathrm{g}}^{1,1}\)

No. of unpaired e : 5

So \(\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}\) and \(\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}\) have same magnetic moment (spin only).

More the nuclear charge on ion, electrons are more strongly attracted, and thus size decreases. Therefore the correct order is:Na⁺ > Mg²⁺> Al³⁺> Si⁴⁺. Thus, the smallest size is Si⁴⁺.

The rate of a chemical reaction is the change in concentration over the change in time and is a metric of the "speed" at which a chemical reactions occurs and can be defined in terms of two observables:

1. The Rate of Disappearance of Reactants: \(\frac{[-\Delta \text { Reactants] }}{\Delta \text { time }}\)

2. The Rate of Formation of Products: \(\frac{[\Delta \text { Products }]}{\Delta \text { time }}\)

Thus, the units for the rate is Molarity per Seconds \(\left(\frac{M}{s}\right)\). or Moles per litre per second \(\left(\right.\)moles \(\left.{ }^{-1} s ^{-1}\right)\).

Interstitial compounds have high melting points in comparison to pure metals. They are very hard and retain metallic conductivity. They are chemically inert i.e., not reactive.

• Interstitial compounds are those which are formed when small atoms like H, C, N, B etc. are trapped inside the crystal lattice of metals.
• They are generally non-stoichiometric and neither typically ionic or covalent in nature.

The enthalpy change of a reaction does not depend on different intermediate reactions.

Enthalpy change is a state function so it does not depend on the path taken by the reaction. it depends only on the difference of final and initial values of enthalpy change.

The IUPAC name of neopentane is 2, 2-dimethylpropane.

Vermilion is the common name of Mercury Sulfide.

• It is a chemical compound composed of the chemical elements mercury and sulfur.
• The chemical formula of Mercury sulfide is HgS.
• It is dimorphic with two crystal forms:
• Red cinnabar
• Black metacinnabar

The chelating ligand is a ligand that has two donor atoms to bind a single metal on. Glycinato has 2 oxygen atoms as donor atoms. It is a chelating igand. Oxalato has 2 oxygen atoms as donor atoms. It is a chelating igand. Thiosulphato \(\left(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\right)\) is an ambidentate ligand. not a chelating ligand.

The oxidation of toluene \(\left(\mathrm{C}_{6} \mathrm{H}_{3} \mathrm{CH}_{3}\right)\) with chromyl chloride \(\left(\mathrm{CRO}_{2} \mathrm{Cl}_{2}\right)\) in \(\mathrm{CCl}_{4}\) or \(C S_{2}\) to give benzaldehyde is called Etard reaction.

In this reaction, the chromyl chloride first forms a brown complex, which is seperated and then decomposed with \(\mathrm{H}_{2} \mathrm{O}\) to give benzaldehyde \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHO}\right)\).

The solvent used is carbon disulfide. Chromium chloride converts the methyl group to a chromium complex. This complex on acid hydrolysis gives the corresponding aldehyde.

\(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{3} \stackrel{\mathrm{CrO}_{2} \mathrm{Cl}_{2}}{\longrightarrow} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}\left(\mathrm{OCrOHCl}_{2}\right)_{2} \stackrel{\mathrm{H}_{2} \mathrm{O}}{\longrightarrow} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHO}\)