BITSAT Chemistry Test - 5 - JEE MCQ

The bond length is inversely proportional to the bond order. 
In the given compound:

Due to resonance, there is a delocalisation of pi-electrons so 'a' acquires a double bond character. So, the bond length of 'a' decreases while the bond order increases, i.e., the bond order of double bond is 2
2. Due to the partial double bond, 'a' has a bond order which is greater than one.
'b' is a single bond between oxygen and carbon. Bond order of a single bond is 1.

Bond length is inversely proportional to the bond order. When more electrons participate in a bond formation, the bond becomes shorter.
By approximation, the bond lengths of two different atoms are the sum of the two individual covalent radii. ​​​
Greater the s− s - character in the orbital, stronger and shorter will be the bond.
The bond length of carbon-carbon triple bond is shorter, due to more % s− character (50%)  than carbon-carbon double bond with % s− character as 33% .
Bond length of −C=C− is shorter than resonance bond in benzene, which has a bond length, between single bond and double bond and is shorter than carbon-carbon single bond, which has 25% s−  - character. Therefore,C₂H₆>C₆H₆>C₂H₄>C₂H₂

Electromeric effect is the complete shifting of π electrons present between two atoms in the presence of an attacking reagent. This effect is shown by compounds which have multiple bonds like:C=C, C≡C, N=N, C=O, C≡N etc.
It means π-bond is involved in this effect.
Since this effect takes place in the presence of an attacking reagent, it is a temporary effect. It is of two types i.e., + E and - E.
Positive Electromeric effect

Negative Electromeric effect

Geometrical isomerism is shown by compounds having at least one C = C and the two groups attached to each carbon must be different.

So, 1,1-dichloro-1-pentene does not show geometrical isomerism due to the presence of two similar
Cl atoms on the same C-atom.

In case of kjeldahl's method the percentage of  N₂ is then calculated from the amount of NH₃

CH₃NH₂ and CH₃OH are nucleophiles, CH₃−Cl is an electrophile. But  is a nucleophile due to the presence of a lone pair of electrons on N and is an electrophile due to the presence of a partial positive charge on C.

Mesomeric effect: It is the movement of the pi -electrons in the conjugated system.
Conjugated system (for carbocations):- Means alternate vacant p- orbital and double bonds. 
The carbocation present adjacent to the oxygen atom is stabilised by the +M effect of the oxygen or any other hetero atom (which has a lone pair), and involves donation of lone pair of electrons to the vacant p− orbital of the carbon.
In the option (2) , the vacant p- orbital of the carbon atom and the lone pair of oxygen are not adjacent to each other. 
In option (1) the vacant p- orbital of the carbon atom and the lone pair of oxygen are adjacent to each other, so, the positive charge is stabilised by + M effect. 
In option (3), the​​​​​ vacant p- orbital of the carbon atom and the lone pair of nitrogen are adjacent to each other, so, the positive charge is stabilised by + M effect.
In option (4), the positive charge is stabilised by +M effect of the three phenyl groups. 

Here, in this case, we have to determine the stability of alkoxides. 
Here C  is more stable because in this case -ve charge on O is in conjugation with a double bond and NO₂ group​. So there is an extra double bond in conjugation.
In case of B, -ve charge on oxygen is in conjugation with a double bond and not with NO2_​ group.
In case of A, there is no double bond so, no conjugation possible. So, least stable.
So, stability order is C>B>A

Resonance and inductive effect decide stability of carbocations.

∴  Correct order of stability is
(II) <(I) <(III) <(IV)

Let in the given mix. Mass of CO = x gm

I₂O₅+5CO ⟶ 5CO₂+I₂
1 mole I₂ is produced from 5 moles of CO
=0.01 mole I₂ is produced from 0.05 moles of CO.

BaCl₂ is the limiting reagent. So, the amount of BaSO₄ will correspond to the amount of BaCl₂.

As first oxide is MO₂
Let atomic mass of M = x

Or 0.5 x x + 16 = 32
Or 0.5 x = 16
x=32
∴ At. Mass of metal  m = 32
Let formula of second oxide is M₂O_n

Therefore, formula of second oxide = M₂O₆
 Or =MO₃

Given mass of O2=2 g at 0℃ and 760 mm Hg
32 g of O2 = 22.4 L at STP
∴ 2g of O2 = x 2 = 1.4L

One mole of any element contains NA entities.
Number of atoms in
Number of atoms in

In CS₂
C : S mass ratio is 15.79 : 84.21
15.79 parts of carbon combine with sulphur = 84.21
∴ 27.27 parts of carbon will combine with


Hence, ratio of S : O is 145.434:72.73 ie, 2 : 1
In SO2, the ratio of S : O is 1 : 1
Since, the ratio of S : O is a simple whole number ratio,
Therefore, law of reciprocal proportions is proved.

Divide the no. of moles of each reactant by stoichiometric co - efficient in the balanced equation. Which ever provides least quotient is limiting reagent.

i.e., B is limiting reagent & product side calculations based on this
 4 moles of B → 3 moles of C
6 moles B → 3/4 x 6 = 4.5 moles of C.

The balanced chemical reaction is
3BaCl₂+2Na₃PO₄→Ba₃(PO₄)₂+6NaCl
According to the reaction,
3 mole BaCl₂ reacts with 2 mole Na₃ (PO4)₂
Thus, 2 mole BaCl₂ will requiremole Na₃ PO₄
Since sodium phosphate is taken in less quantity then the required quantity, here, Na₃PO₄ is the limiting reactant. The stoichiometric calculations are done based on the limiting reagent.
According to the reaction.
2 mole Na₃ PO₄ gives 1 mole Ba₃(PO₄)₂
So,1 mole Na₃PO₄ will give 0.5 mole Ba₃(PO₄)₂.

Hence, the simplest molecular formula(empirical formula) C₅H₇N
The empirical Mass = 81
Now, molecular formula can be calculated using the ratio of molecular mass and empirical mass.
The ratio of molecular mass and empirical mass = 2 : 1
Hence, the molecular formula of the compound is C₁₀H₁₄N₂.

(i) 1 mole of ethane, C₂H₆ contains 2 moles of carbon atoms.
∴ 3 moles of ethane, C₂H₆ will contain 2 × 3 = 6 moles of C-atoms.
(ii) 1 mole ethane, C₂H₆ will contains 6 moles of H-atoms.
∴ 3 moles of ethane, C₂H₆ will contain 3 × 6 = 18 moles of H-atoms.
(iii) 1 mole of ethane, C₂H₆ = 6.022 × 10²³ molecules of ethane.
∴ 3 moles of ethane,C₂H₆= 3 × 6.022 × 10²³ = 18.066 × 10²³ molecules of ethane.

Correct option is D.
An alpha particle is obtained by removing 2 electrons from a helium atom. So, an alpha particle is doubly - charged helium ion.

The number of atoms in a simple cubic unit cell is one because contribution of atom at eight corners of the cube = 

In nuclear reactors, the control rods are made up of cadmium because cadmium rod absorbs excess neutrons.

Solder is an alloy of lead. It is made up of 33% of lead and 67% of Tin.

P orbitals take a maximum of 6 electrons. 2px², 2py², 2pz². The p orbitals are along 3 perpendicular axis. So p orbitals can accommodate a maximum of 6 electrons.

In SO₄^2-, PO₄^3- and BF₄^- all the central atoms are in sp³ - hybridisation state. Hence they are isostructural species (tetrahedral).

The aqua-regia is a mixture of one part of cone. HN03 and three parts of cone. HCI.

Absolute zero is the temperature of which the velocities of the gas molecules reduce to zero. It means molecular motion ceases at absolute zero.

In sunlight, sulphurdioxide and chlorine reacts as follows :

The surface water contains suspended impurities.

Avogadro’s number is 6.023 x 10²³ which is equal to molecules present in gram molecular mass.