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The number of sigma $(\sigma)$ and pi$(\pi)$ bonds in pent-2-en-4-yne is:

• Option 1)

10 $\sigma$ bonds and 3 $\pi$ bonds

• Option 2)

$\sigma$ bonds and 5 $\pi$ bonds

• Option 3)

11 $\sigma$ bonds and 2 $\pi$ bonds

• Option 4)

13 $\sigma$ bonds and no $\pi$ bonds

Option 1 Structure of pent-2-en-4-yne Here there are 10  bonds  and  bonds And 3  bonds   and Option 1)10  bonds and 3  bondsOption 2)8  bonds and 5  bondsOption 3)11  bonds and 2  bondsOption 4)13  bonds and no  bonds

An alkane "A" on reaction with O3 and Zn-H2O gives propane and ethanal in equimolar ratio. Addition of HCl to Alkane "A" gives "B" as the major product. The structure of the product "B" is

• Option 1)

• Option 2)

• Option 3)

• Option 4)

This oxidation of alkene occurs at the double bond. Thus, propane & ethanol must combine to give the alkene             (Propanone)                                                                                                       Now this reacts with HCl to form B as follows:                                                                         ...

Which of the following acts as a circuit protection device ?

• Option 1)

conductor

• Option 2)

inductor

• Option 3)

switch

• Option 4)

fuse

fuse Option 1)conductor Option 2)inductor Option 3)switch Option 4)fuse

In which of the following devices , the eddy current  effect is not used ?

• Option 1)

induction furnace

• Option 2)

magnetic braking in train

• Option 3)

electromagnet

• Option 4)

electric heater

electric heater Option 1)induction furnace Option 2)magnetic braking in train Option 3)electromagnet Option 4)electric heater

Match the Xenon compound in Column-I with its structure in Column-II and assign the correct code:

Column-I                        Column-II

(a) XeF4                            (i) pyramidal

(b) XeF6                            (ii) Square planar

(c) XeOF4                        (iii) distorted octahedral

(d) XeO3                          (iv) square pyramidal

Code:

(a)     (b)     (c)     (d)

• Option 1)

(i)     (ii)     (iii)     (iv)

• Option 2)

(ii)      (iii)     (iv)     (i)

• Option 3)

(ii)     (iii)     (i)     (iv)

• Option 4)

(iii)     (iv)     (i)     (ii)

XeF4  it is square planar. XeF6  it is distorted octahedral because of the lone pair present. XeOF4  Its is square pyramidal & its hybridization is sp3d2 XeO3  Its hybridization is sp3 & its shape is pyramidal.  (a) - (ii), (b) - (iii), (c) - (iv), (d) - (i) Option 2 is correct.Option 1)(i)     (ii)     (iii)     (iv)Option 2)(ii)      (iii)     (iv)     (i)Option 3)(ii)     (iii)     (i)   ...

Pick the wrong answer in the context with rainbow

• Option 1)

When the light rays undergo two internal  reflection in a water drop , a secondary rainbow is formed

• Option 2)

The order of colors is reversed in the secondary rainbow

• Option 3)

An observer can see a rainbow when his front is towards the sun

• Option 4)

Rainbow is a combined effect of dispersion , refreaction and reflection of sunlight

Rainbow can seen when back is towards the Sun.Option 1)When the light rays undergo two internal  reflection in a water drop , a secondary rainbow is formed Option 2)The order of colors is reversed in the secondary rainbow Option 3)An observer can see a rainbow when his front is towards the sun Option 4)Rainbow is a combined effect of dispersion , refreaction and reflection of sunlight

A particle moving with velocity $\vec V$ is acted by three forces shown by the vector triangle PQR . The velocity of the particle will :

• Option 1)

increase

• Option 2)

derease

• Option 3)

remain constant

• Option 4)

change acording to the smallest force QR

Option 1) increase  Option 2) derease  Option 3) remain constant  Option 4) change acording to the smallest force QR

Which mixture of the solution will lead to the formation of negatively charged colloidal $[AgI]I^-$ sol. ?

• Option 1)

50 mL of 1 M $AgNO_3$ + 50 mL of 1.5 M $KI$

• Option 2)

50 mL of 1 M $AgNO_3$ + 50 mL of 2 M $KI$

• Option 3)

50 mL of 2 M $AgNO_3$ + 50 mL of 1.5 M $KI$

• Option 4)

50 mL of 0.1 M $AgNO_3$ + 50 mL of 0.1 M $KI$

Option 2                     Negatively charged colloidal  need 2 mol of   Option 1)50 mL of 1 M  + 50 mL of 1.5 M Option 2)50 mL of 1 M  + 50 mL of 2 M Option 3)50 mL of 2 M  + 50 mL of 1.5 M Option 4)50 mL of 0.1 M  + 50 mL of 0.1 M

Which of the following is incorrect statement?

• Option 1)

PbF4 is covalent in nature

• Option 2)

SiCl4 is easily hydrolysed

• Option 3)

GeX4 (X = F, Cl, Br, I) is more stable than GeX2

• Option 4)

SnF4 is ionic in nature

(i) PbF4 is not covalent in nature. Because Pb atom is very & F atom is very small, so according to Fajan's rule, PbF4 is not covalent. (ii) SiCl4 is not a very stable compound because of steric hindrance thus can be hydrolysed very easily (iii) GeX4 is more stable than GeX2 as in GeX4 all the orbitals are fully filled.  (iv) SnF4 is ionic in nature. Because F atom is very small & Sn atom is...

Two parallel infinite line charges with linear charges densities $+ \lambda C/ m \: \: and \: \: - \lambda C/ m$ are placed at a distance  of 2R in free space . what is the electric field mid way between the two line charges ?

• Option 1)

zero

• Option 2)

$\frac{2 \lambda }{\pi \epsilon _0 R} N/C$

• Option 3)

$\frac{ \lambda }{\pi \epsilon _0 R} N/C$

• Option 4)

$\frac{\lambda }{2\pi \epsilon _0 R} N/C$

Option 1) zero Option 2) Option 3) Option 4)

What is the correct electronic configuration of the central atom in $K_6[Fe(CN)_6]$ based on crystal field theory?

• Option 1)

$t_{2g}^4e_{g}^2$

• Option 2)

$t_{2g}^6e_{g}^0$

• Option 3)

$e^3t_2^3$

• Option 4)

$e^4t_2^2$

Splitting occurs as follows for octahedral For Fe2+ (d6), electron distribution occurs as follows:  is the correct distribution. Option 2 is correct.  Option 1)Option 2)Option 3)Option 4)

Two similar thin equiconvex lenses , of focal length f each , are kept  coaxially in contact  with each other such that the focal length of the combination is $F_1$ . When the space between the two lenses is filled  with glycerin  then the equivalent  focal length is $F_2$ . The ratio $F_1 : F_2$

• Option 1)

2 : 1

• Option 2)

1 : 2

• Option 3)

2 : 3

• Option 4)

3 : 4

Option 1)2 : 1 Option 2)1 : 2 Option 3)2 : 3 Option 4)3 : 4

The non-essential amino acid among the following is:

• Option 1)

valine

• Option 2)

leucine

• Option 3)

alanine

• Option 4)

lysine

Non essential amino acids are those which produce in the human body itself. Essential amino acids are those which do not produce in the human body & we have to take them from outside sources. Alanine is the non-essential amino acid. Option 3 is correct.Option 1)valineOption 2)leucineOption 3)alanineOption 4)lysine

For an ideal solution, the correct option is:

• Option 1)

$\Delta_{mix}S = 0$ at constant T and P

• Option 2)

$\Delta_{mix}V \neq 0$ at constant T and P

• Option 3)

$\Delta_{mix}H = 0$ at constant T and P

• Option 4)

$\Delta_{mix}G = 0$ at constant T and P

For an ideal solution, the Gibbs free energy is always negative, thus their mixing is always spontaneous. The volume change of the mixing of ideal solutions is zero. An ideal solution is one in which the enthalpy of the solution is zero  at constant T and P Option 3 is correct.Option 1) at constant T and POption 2) at constant T and POption 3)  at constant T and POption 4) at constant T and P

For the cell reaction

$2Fe^{3+}(aq)+ 2I^{-}(aq) \rightarrow 2Fe^{2+}(aq) + I_2(aq)$

$E^0_{cell} = 0.24V$ at 298K. The standard Gibbs energy $(\Delta_r G^0)$ of the cell reaction is:

[Given that Faraday constant F = 96500 C mol-1]

• Option 1)

-46.32 kJ mol-1

• Option 2)

-23.16 kJ mol-1

• Option 3)

46.32 kJ mol-1

• Option 4)

23.16 kJ mol-1

We know that       Option 1 is correctOption 1)-46.32 kJ mol-1Option 2)-23.16 kJ mol-1Option 3)46.32 kJ mol-1Option 4)23.16 kJ mol-1

Enzymes that utilize ATP in phosphate transfer require an alkaline earth metal (M) as the cofactor M is:

• Option 1)

Be

• Option 2)

Mg

• Option 3)

Ca

• Option 4)

Sr

Option 2 Mg is the alkaline earth metal as the co-factor. (FACT)Option 1)BeOption 2)MgOption 3)CaOption 4)Sr

A compound is formed by cation C and anion A. The Anions form hexagonal close packed lattice and the cations occupy 75% of the octahedral void. The formula of the compound is

• Option 1)

$C_2A_3$

• Option 2)

$C_3A_2$

• Option 3)

$C_3A_4$

• Option 4)

$C_4A_3$

Option 3 Anions: No of ions per unit in HCP = 6 Cations: No of octahedral voids in HCP = 6                         So, 75% of 6 = 4.5     Option 1)Option 2)Option 3)Option 4)

A gas at 350K and 15 bar has molar volume 20 percent smaller than that for an ideal gas under the same conditions. The  correct  option about the gas and its compressibility factor (Z) is:

• Option 1)

Z>1 and attractive forces are dominant

• Option 2)

Z>1 and repulsive forces are dominant

• Option 3)

Z<1 and attractive forces are dominant

• Option 4)

Z<1 and repulsive forces are dominant

Option 3          Molar volume of gas is 20% smaller than that of the ideal gas. Hence Z < 1 and attractive forces are dominant. When Z < 1: Gas shows negative deviation and is more compressible than expected from ideal behaviour.  Z < 1 corresponds to attractive forces between molecules.Option 1)Z>1 and attractive forces are dominantOption 2)Z>1 and repulsive forces are dominantOption 3)Z<1...

• Option 1)

nylon-6,6

• Option 2)

nylon 2 - nylon 6

• Option 3)

nylon-6

• Option 4)

Buna-S

Option 2  Biodegradable polymer is Nylon-2-Nylon-6 monomers: Glycine, aminocaproic acid Option 1)nylon-6,6Option 2)nylon 2 - nylon 6Option 3)nylon-6Option 4)Buna-S

For the chemical reaction

$N_2(g) + 3H_2(g)\leftrightharpoons 2NH_3(g)$

the correct option is:

• Option 1)

$-\frac{1}{3}\frac{d[H_2]}{dt} = -\frac{1}{2}\frac{d[NH_3]}{dt}$

• Option 2)

$-\frac{d[N_2]}{dt} = {2}\frac{d[NH_3]}{dt}$

• Option 3)

$-\frac{d[N_2]}{dt} = \frac{1}{2}\frac{d[NH_3]}{dt}$

• Option 4)

${3}\frac{d[H_2]}{dt} ={2}\frac{d[NH_3]}{dt}$

For a chemical reaction: The rate is always given as:  for the reaction: , Thus,  Option 3 is correct.Option 1)Option 2)Option 3)Option 4)
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