Chemistry NCERT Exemplar Solutions Class 12th Chapter Seven The p-Block Elements

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Oxygen is more electronegative than chlorine, the negative charge on chlorine disperses from ClO⁻ to ClO₄⁻ ion as the number of oxygen atoms linked to chlorine grows. As a result, ion stability will improve in the following order:

As the stability of the conjugate base rises, the acidic strength of the corresponding acid increases in the following order:

HClO < HClO₂ < HClO₃ < HClO₄

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Three P—Cl bonds are arranged in one plane at a 1200 angle. These bonds are known as equatorial bonds because they are formed between two points. The remaining two P—Cl bonds, one above the equatorial plane and the other below it, form  90° angle with the plane. Axial bonds are the name for these types of bonds. Because axial bond pairs are subjected to greater repulsive interaction than equatorial bond pairs, axial bonds are slightly longer and hence slightly weaker than equatorial bonds, making the PCl₅ molecule more reactive.

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NO₂ exists as a monomer with one unpaired electron in the gaseous state, but it dimerises to N₂O₄ in the solid state, leaving no unpaired electron, making the solid form diamagnetic.

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Chlorine has vacant d-orbitals, which become excited upon bonding when electrons from the 3p-orbital are promoted to the 3d-orbital, giving it a covalency of three.

Due to the lack of unoccupied d-orbitals in the second energy shell, fluorine cannot expand its octet. As a result, it can only have one covalency.

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Oxygen is more electronegative than sulphur, the bond angle of H₂O is greater, and the bod pair electron of an OH bond will be closer to oxygen, causing bond-pair repulsion between bond pairs of two OH bonds

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Six F atoms can be accommodated around sulphur due to fluorine's tiny size, however the chloride ion is comparably bigger, resulting in interatomic repulsion.

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A is PCl₅ (yellowish white powder)

P₄ + 10Cl₂→4PCl₅

B is PCl₃ (colourless oily liquid)

Hydrolysis of the following product

PCl₃+3H₂O→H3PO₃+3HCl

PCl₅+4H₂O→H3PO₄+5HCl

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NO₃⁻ + 3Fe₂⁺ + 4H⁺→ NO⁺ + 3Fe₃⁺ + 2H₂O

[Fe (H₂O)₆]²⁺ + NO → [Fe (H₂O)₅ (NO)]²⁺ + H₂O ( brown ring )

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Due its disintegration into oxygen results in the release of heat ΔH is negative) and an increase in entropy (ΔS is positive), ozone is thermodynamically unstable in comparison to oxygen. These two reactions combine to produce a substantial negative Gibbs energy change (ΔG) for its conversion to oxygen.

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P₄O₆ + 6H₂O→4H₃PO₃  (i)

Next is neutralization

4 times H₃PO₃ + 2NaOH→Na₂HPO₃ + 2H₂O  (ii)

On addition of 2 equations

P₄O₆+8NaOH→4Na₂HPO₃+2H₂O

P₄O₆ ( mol.mass )= (4×31+16×6)=220

Number of moles of P₄O₆ = $\mathit{?}/\mathit{V}$    = $\mathit{?}$

The product formed is neutralized 8 moles of NaOH

P₄O₆ = 8× $\mathit{?}$  = $\mathit{?}$  molNaOH

Molarity of NaOH in 1 litre  =  0.1M

 Molarity = $\frac{\mathrm{M}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{M}\mathrm{a}\mathrm{s}\mathrm{s}}{\mathrm{G}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{n}\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}}$

 Volume = $\frac{1.1}{220}$  

= $\frac{1.1}{220}$ × $\frac{8.8}{220}$ = 0.4 L

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P₄+6Cl₂→4PCl₃…  (i)

PCl₃+3H₂O→H₃PO₃+3HCl ×4

On adding eq. (i) and (ii)

P₄+6Cl₂+12H₂O→4H₃PO₃+12HCl

1 mol of white phosphorus produces 12 mol of HCl

62 g of white phosphorus has been taken which is equivalent to $\frac{\mathrm{N}\mathrm{o}.\mathrm{o}\mathrm{f}\mathrm{}\mathrm{M}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{ }\mathrm{ }}{\mathrm{V}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}\mathrm{}\mathrm{i}\mathrm{n}\mathrm{}\mathrm{l}\mathrm{i}\mathrm{t}\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{ }}$ = $\frac{\mathrm{N}\mathrm{o}.\mathrm{o}\mathrm{f}\mathrm{}\mathrm{M}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{ }\mathrm{ }}{\mathrm{M}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{i}\mathrm{t}\mathrm{y}}$ mol. Therefore 6 molHCl will be formed.

Mass of 6 molHCl=6×36.5=219.0g HCl

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Three oxoacids are

(a) HNO₂, Nitrous acid

(b) HNO₃, Nitric acid

(c) Hyponitrous acid, H₂N₂O₂

3HNO₂ + H₂O + 2NO

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On interaction with P₄O₁₀ forms an oxide of nitrogen,  N₂O₅, and metaphosphoric acid, 3, nitric acid is formed. HPO₃

HNO₃ + P₄O₁₀→4HPO₃ + 2N₂O₅

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The crystal structure of red phosphorus features a sophisticated network of bonding, whereas white phosphorus is made up of P₄ molecules. To avoid natural combustion, white phosphorus must be stored in water, but red phosphorus is stable in air.

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When nitric acid is mixed with copper metal, it produces distinct oxidation products.

3Cu + 8HNO₃ (dil.)→3Cu (NO₃)₂ + 2NO + 4H₂O

Cu + 4HNO₃ (Conc)→3Cu (NO₃)₂ + 2NO₂ + 2H₂O

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PCl₅ + 2Ag→2AgCl + PCl₃

AgCl + 2NH₃ (aq)→ [Ag (NH₃)₂] + Cl^- 

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The following reaction with silver nitrate demonstrates phosphinic acid reducing behaviour:

4AgNO₃ + 2H₂O + H₃PO₂→4Ag + 4HNO₃ + H₃PO₄

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4NH₃ + 5O₂ 4NO + 6H₂O

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‘A’ is S₈ ‘B’ is SO₂ gas

 S₈ + 8O₂ 8SO₂

2MnO₄ ^– + 5SO₂ + 2H₂O → 5 SO₄^2– + 4H⁺ + 2Mn²⁺

(violet)                                                              (colourless)

2Fe³⁺ + SO₂ + 2H₂O → 2Fe²⁺ + SO₄^2– + 4H⁺

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Here, ‘A’ is NO₂ (Nitrogen dioxide)

‘B’ is N₂O₄ (dinitrogen tetraoxide)

‘C’ is N₂O₃ (dinitrogen trioxide)

A brown gas is produced when lead nitrate (II) is heated

2Pb (NO₃)₂ 2PbO  +  4NO₂ + O₂

2NO₂ N₂O₄

2NO  + N₂O₄ 2N₂O₃ 

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Correct option is (i)

CCl₃NO₂ stands for chloropicrin,  which is known as tear gas. Cl is the element of the highest electron gain enthalpy. Sulphur is the element belonging to 16th group,  also known as chalcogens.

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Correct option is (ii)

This is a matching answer type question as classified in NCERT Exemplar

Correct option is (iii)

This is a assertion and reason answer type question as classified in NCERT Exemplar

Correct option is (ii)

Because H₂SO₄ is a stronger oxidising agent and HI is a stronger reducing agent,  H₂SO₄ oxidizes and is reduced to SO₂.

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Correct option is (i)

The rhombic and the monoclinic forms of sulphur exist as S8 . Oxygen forms a p\pi   -  p\pi  multiple bond due to its small size and short bond length,  but Pπ- Pπ bonding is not possible because of sulphur's larger atomic size.

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Correct option is (i)

When NaCl reacts with concentrated H₂SO₄,  it produces colourless fumes with a pungent smell,  but when MnO₂ is added, the fumes turn greenish yellow. MnO₂ oxidizes HCl to chlorine gas,  which then turns greenish yellow.

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Correct option is (i)Here,  the S atom is surrounded by 6F atoms,  so it's complected for H₂O to attack SF₆,  but in case of SF₄,  it's surrounded by 4F atoms,  so it's to attack H₂O.

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Correct option is (iii) H₃PO₄

When it reacts with a metal atom, it can produce three different forms of salts, referred known as the three series of salts.

H₃PO₄ + Na→NaH₂PO₄

NaH₂PO₄ + Na→Na₂HPO₄

Na₂HPO₄ + Na→Na₃PO₄

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Correct option is (i)

We get N₂ in both situations when we heat ammonium dichromate and barium azide separately.

(NH₄)2Cr₂O₇→Cr₂O₃ + 4H₂O + N₂

Ba (N₃)₂→Ba + 3N₂.

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Correct option is (i)

Peroxy linkage refers to the presence of a bond between oxygen and oxygen (O-O) in a molecule.

Only H₂S₂O₆ and H₂S₂O₇ of the sulphur oxoacids above have a peroxy linkage.

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Correct option is (iii)

Bartlett noticed that Xenon's first ionisation potential is nearly identical to that of oxygen and used Born-Haber calculations to predict the presence of a stable compound: Xe⁺ Pt F₆^- 

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Correct option is (ii)

BrO₂^- (35+2×8+1=52) and BrF₂⁺  (35+2×9−1=52)

52 electrons are present in both 

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Correct option is (ii), (iii)

MI  >  MBr  >  MCl  >  MF is an iconic character of metal halide

F₂ >  Cl₂ >  Br₂ > I₂ is a bond dissociation enthalpy

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Correct option is (i) and (iii)

SO₂ is employed as an antichlor, disinfectant, and preservative in the bleaching of wool and silk.

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Correct option is (i) and (iv)

(i) As₂O₃ < SiO₂ < P₂O₃ < SO₂ is the order of acid

(ii) AsH₃ <  PH₃ < NH₃ is the correct or enthalpy of vapourization

(iii) S < O < F < Cl is the correct order of more negative gain enthalpy

(iv)  H₂O > H₂Se > H₂Te is the order of thermal stability

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Correct option is (i)

Oxidation state of S=+6

(iii) Iron oxide with K₂O and Al₂O₃ is used to increase the rate of attainment of equilibrium in Haber's process.

(iv) Change in enthalpy is negative for the preparation of SO₃ by catalytic oxidation of SO₂.

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Correct option is (i) and (ii)

(i) The modest dispersion force causes attraction in noble gases.

(ii) In nature, xenon fluorides are reactive.

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Correct option (iii)

If an electron pair is donated to a vacant orbital on one atom and a lone pair of electrons on the other, the bonding is designated pπ-dπ depending on the orbital to which the electron pair is donated and the orbital from which the electron pair is donated.

Because their valence shells lack d-orbitals, nitrogen, carbon, and boron cannot form a pi bond. Phosphorus, on the other hand, can produce pi bonds.

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Correct option (i)

The compound has the following number of electrons:

NO₃^-  =  32e^- 

CO₃^2-  =  32e^- 

ClO₃^-  =  42e^- 

SO₃ ^-  =  42e^- 

Hence,  NO₃^- and CO₃^2-  are isoelectronic

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Correct option (i)

Fluorine is the most reactive element, forming hydrogen fluoride when it combines with hydrogen gas.

Due to the shorter bond length between the two atoms, it takes a lot of energy to break the link between hydrogen and fluorine.

As a result, as we move down the group, the bond dissociation energy drops.

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Correct option is (iv)

The E-H bond dissociation energy of SbH₃ is the lowest among the aforementioned compounds, it easily releases H, making it a strong reducing agent. The greater the lowering agent, the weaker the E-H bond.

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Correct option is (iii)

When heated with concentrated NaOH in an inert CO₂ environment. It formsPH₃

P₄ + 3NaOH + 3H₂O→PH₃ + 3NaH₂PO₂

As we progress from the top to the bottom in a group, the basic character of hydrides reduces PH₃ is basic than NH₃.

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Correct option is (iii)

H₃PO₂ has one O-H group and two P-H bonds

The existence of a P-H link gives phosphorus oxyacids their reducing characteristics. It has a strong proclivity for releasing protons. As a result, it demonstrates a decreasing nature.

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Correct option is (ii)

On heating, lead nitrate decomposes into lead monoxide, nitrogen dioxide, and oxygen.

2Pb (NO₃)₂→2PbO + 4NO₂ + O₂.

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Correct option is (i)

The single N-N bond is weak because of high interelectronic repulsion of the non-bonding electrons, owing to the small bond length. As a result the catenation tendency is weaker in nitrogen that is why it does not show allotropy

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Correct option is (ii)

The valence electrons of nitrogen are two 2s and three 2p. Nitrogen can make three covalent bonds by sharing its three 2p electrons. Consider the NH₃ molecule. The nitrogen atom, however, still contains a lone pair of electrons in the 2s orbital. It can take one bond by transferring these two electrons from the lone pair. Consider the NH₄⁺ molecule.

As a result, nitrogen can create four different bonds.

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Correct option is (i)

The N-N sigma bond (single bond) is weaker than the P-P sigma bond (single bond) due to the smaller size of nitrogen which increases the interelectronic repulsions and thus making the bond weaker. 

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Correct option is (i)

The 'Brown Ring Test' is an identification and conformation test of the nitrate ion  (NO₃^- ) using conc. H₂SO₄ and newly produced FeSO₄. Place the NO₃^-  containing sample in a test tube. Then, with gentle shaking, add conc H₂SO₄ dropwise, heat it briefly on a Bunsen burner, and add freshly made greenish FeSO4 Then, due to the creation of brown coloured  [Fe (H₂O)₅ (NO)]SO₄ complex, a brown ring forms in that test tube.

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Correct option is (ii)

Bismuth is the sole element that has an inert pair effect. Only generates trihalides and has a +3 oxidation state. Florine, on the other hand, is small and has a high electronegativity.

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Correct option is (i)

The catalytic oxidation of ammonia produces NO gas, which is used to make HNO₃. 4 moles of NH₃ created 4 moles of NO in the equation below. As a result, the moles of NO produced by oxidising two moles of NH₃ will be two moles.

4NH₃ + 5O₂→4NO + 6H₂O.

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Correct option is (iii)

Na (H₂)PO₂

1+ (2x+1)+ x +2 (−2)=0

x−1=0

x=1

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Correct option is (iii)

The C-atom in the CO₃^2-  ion undergoes sp2 hybridization. BF₄, NH₄⁺  and SO₄^2-  have a tetrahedral structure, whereas it has a triangular planar structure.

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Correct option is (iii)

C+2H₂SO₄→CO₂+2SO₂+2H₂O

Conc. H₂SO₄ oxidises C to produce two gaseous products.

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Correct option is (iv)

MnO₂ ( Black )+4HCl→MnCl₂+2H₂O+Cl₂

Greenish yellow color

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Correct option is (iii)

The ability of a substance to be decreased is known as its reduction potential. As the substance's reduction potential rises, so does its power to reduce. It signifies the chemical's oxidising power (the ability of the material to cause other substances to lose electrons, which is known as oxidising power) grows.

As a result, the oxidising power is listed in decreasing order BrO₄^-  >  IO₄^-  >  ClO₄^-  

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Correct option is (i) and (iii)

6NaOH+3Cl₂ → 5NaCl+NaClO₃+3H₂O

Chlorine gas oxidation number ranges from 0 to –1 to 0 to +5.

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Correct option is (ii) and (iv)

Tetrahedral geometry characterises the structure of a white phosphorus molecule.

Phosphorus has a valency of 5, and there are four phosphorus atoms in total. As a result, there will be a total of 4×5=20 valence electrons.

As a result, each P atom will have one lone pair of electrons, and each covalent bond (two electrons) will be established.

In a molecule of white phosphorus, there will be four lone pairs of electrons and six P-P single bonds in total.

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Correct option is (i)

Due to its high electronegativity and tiny size, F is the only interhalogen with weaker X-X’ bonds and no X-X bonds.

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Correct option is (iii) and (iv)

The P₄ molecule in white phosphorus has an angular strain, white phosphorus is very reactive.

PCl₅ is ionic in the solid state, with a tetrahedral cation and an octahedral anion.

Siderite – FeCO₃ (ore of iron)

Calamine – ZnCO₃ (ore of zinc)

Malachite – CuCO₃.Cu (OH)₂ (ore of copper)

Cryolite – Na₃AlF₆ (ore of aluminium)

Since SOCl₂ is used to covert aliphatic (R-OH) into chlorides. It will not react with aromatic alcohol

Tritium is radioactive and it decays into He³ during emission of β-radiation

₁T³ → ₂He³ + _-1e⁰

$FeC{l}_3\underset{}{\overset{hydrolysis}{\to }}Fe{\left(OH\right)}_3\downarrow \underset{Adsorption}{\overset{F{e}^{3+}}{\to }}Fe{\left(OH\right)}_3|F{e}^{3+}\left(colloidalparticle\right)$

(c) $2C{H}_{3}C{H}_{2}Cl\underset{ether}{\overset{2Na}{\to }}{C}_2{H}_5-C_2{H}_5+2NaCl\left(WurtzReaction\right)$

(d) $2C_6{H}_{5}Cl\underset{ether}{\overset{2Na}{\to }}{C}_6{H}_5-C_6{H}_5+2NaCl\left(WittingReaction\right)$

Ceric ammonium nitrate is used to test alcohol while CHCl₃/alc. KOH is used to test 1° amine

It does not contains 2^nd most abundant element by weight in earth crust because that is Si Calgon

$\underset{}{\overset{Watersoluble}{\to }}2N{a}^{+}{\left[N{a}_4{\left(P{O}_3\right)}_6\right]}^{2-}$

$\underset{}{\overset{C{a}^{2+}}{\to }}2N{a}^{+}{\left[N{a}_{2}Ca{\left(P{O}_3\right)}_6\right]}^{2-}$

Consider the following image

(a) Sucrose  → α-D glucose and β-D fructose

(b) Lactose  → β-D- galactose and β-D glucose

(c) maltose  → α-D glucose and α-D-glucose

$Mn{O}_2+4HCl\to MnC{l}_2+C{l}_2+2H_{2}O$

$\begin{array}{l}C{l}_2+2Kl\to 2KCl+I_2\\ I_2+2N{a}_2{S}_2{O}_3\to 2Nal+N{a}_2{S}_4{O}_6\end{array}$

Here, meq of MnO₂ = meq of Na₂S₄O₆

$\frac{w}{E}\times 1000=M\times n-factor\times V$

$\frac{w}{87/2}\times 1000=0.1\times 1\times 60$

Mass of MnO₂ in sample = 0.261 g

Percentage of MnO₂ in sample = $\frac{0.261}{2}\times 100$  

 = 13.05%

Work function, ${?}_0=6.63\times 10^{-19}J$  

Threshold wavelength  ${\lambda }_0=?$  

Using ;                 ${?}_0=\frac{hc}{{\lambda }_0}$  

${\lambda }_0=\frac{hc}{{?}_0}$  

=300 × 10^-9 m = 300 nm

Hybridization of P in PF₅ is sp³d, so value of y = 1

For expansion in vacuum, workdone, w = 0

For isothermal process,   $\Delta U=0$  

According to first law of thermodynamics,

$\begin{array}{l}\Delta U=q+w\\ q=0\end{array}$  

$P_A^0=50torr$  

$P_B^0=100torr$  

Mole fraction of A in liquid phase,           x_A = 0.3

Mole fraction of B in liquid phase,            x_B = 0.7

Now;     $P_A=P_A^0{x}_A$  

$P_B=P_B^0{x}_B$

= 100 × 0.7 = 70 torr

Mole fraction of B in vapour phase,   $y_B=\frac{P_B}{P_A+P_B}=\frac{70}{85}$  

$\frac{70}{85}=\frac{x}{17}$  

                             X = 14

Using ; k_sp = 108 S⁵

1.1 × 10^-23 = 108 S⁵

S =  ${\left(\frac{110}{108}\times 10^{-25}\right)}^{1/5}M\approx 10^{-5}M$  

Specific conductance,

${\lambda }_m^0=\frac{\kappa }{5\times 1000}S{m}^{2}mo{l}^{-1}$  

= 3 × 10^-3 Sm² mol^-1

So; x = 3

$C{r}_2{O}_7^{2-}+6e^{-}+14H^{+}\to 2C{r}^{3+}+7H_{2}O$  

Here, 6F electricity is required to reduce 1 mol  $C{r}_2{O}_7^{2-}$ to Cr³⁺.

Rate constant, k = 5.5 × 10^-14 s^-1.

$t=\frac{2.303}{k}log\frac{{\left[R\right]}_0}{\left[R\right]}$  

 = $\frac{2.303}{k}log3-(i)$  

$t_{50\mathrm{\%}}=\frac{2.303}{k}log2-(ii)$  

 From (i) & (ii)

$\frac{t_{67\mathrm{\%}}}{t_{50\mathrm{\%}}}=\frac{log3}{log2}$  

= 1.58 t_50%

So;         t_67% is 15.8 × 10^-1 times half life.

X = 16 (the nearest integer)

All metal carbonyls have synergic bonds

Organic compound → AgBr (s)

  0.5 g                    0.40 g

Here; Moles of Br in organic compound = Moles of Br in Ag Br

= Moles of AgBr

=  $\frac{0.40}{188}mol$

Mass of Br in organic compound =  $\frac{0.4}{188}\times 80g$  

$\begin{array}{l}\mathrm{\%}ofBr=\frac{0.4\times 80}{188\times 0.5}\times 100\mathrm{\%}\\ =34\mathrm{\%}\end{array}$