NEET Practice Test - 11 - NEET MCQ

According to lewis's theory, the element which donates lone pair of electrons is basic.

When electrons of lone pair are get involved in resonance, the basicity decreases.

because the electrons of the lone pair are not free to donate.

In the above image, the lone pairs of N of NMe₂ are not involved in resonance due to steric hindrance. Hence, it is more basic.

Hence, the correct answer is option C.

Radius of hydrogen atom (r) = 0.53 Å.

Atomic number of Li (Z) = 3.

Radius of Li²⁺ ion

Vitamin B₁₂ - Pernicious anemia

Vitamin C - Scurvy

Vitamin E - Sterility

The correct relation is b > a because bond ‘a’ has partial double bond characters due to resonance.

According to graph,

E_a(b) = E_a(f) + DH

= 17 + 40

= 57 kJ mol^–1

Bohr's atomic model explains the spectra of single-electron atoms like H-atom. It does not explain the spectra of multi-electron atoms.

So the atoms which are isoelectronic and have the same electronic configuration are well explained by this atomic model.

He⁺, Li²⁺ and tritium ₁H³ have same number of electrons as that of hydrogen atom i.e. 1 while Be²⁺ is formed by losing 2e. So, it is a two-electron system.

⇒ [Cr(SCN) (H₂O)3 (en)] (C2O4) : It can exhibit linkage and geometrical isomerism.

⇒ [ZnBr(CN) (SCN) (NH₃)]^– : It exhibits optical isomerism.

⇒ [Co(BrCl) (H₂O)₄] [Ag(CN)₂] : and [CoCl(CN)(H₂O)₄] [AgBr(CN)] are coordination isomers.

Mole of Au⁺ 25/100 = 1/40 = 0.025

Similarly, moles of Ag⁺ = 0.025

Charge passed = I × t = 1 × 15 × 60 = 900 C

Moles of e^– passed 900/96500 = 0.0093 mol.

Species with higher value of SRP will get deposited first at cathode.

So, only Au will get deposited.

Xe is a noble gas. It has 8 electrons in its valence shell. Now, it is bonded to 4F atoms. 

So, 4 of its electrons are involved in bonding. 

Hence, bond pairs = 4. Remaining 4 are unbonded electrons. So, lone pairs = 4/2 = 2.

At high temp. I.e 400°C substitution occurs in preference to addition.

Due to stability of 2^o− radical, it is easy to synthesize n-butane,

The Wurtz reaction is a coupling reaction in organic chemistry, whereby two alkyl halides are reacted with sodium metal in dry ether solution to form a higher alkane

Vapour pressure will be the same because it doesn't depend on the volume it depends only on Temperature.

Number of H-atoms = 0.0747 × 22 × 6.023 × 10²³

= 9.91 × 10²³

for weak electrolytes

∴  The one with lowest concentration has highest molar conductance.

Hence, option B is correct.

In the trivial system of naming organic compounds, the names of compounds are dependent on the source of the organic compounds or their properties.

If the prefix is iso, then there will be one methyl group on the secondary carbon of the parent chain. And if the prefix is neo, then there will be two methyl groups on the secondary carbon of the parent chain.

Also, heptyl is there that means there are a total of 7C present and the alcohol is primary.

Hence, Neo-heptyl alcohol is correctly represented as:

1 ml of 0.1 M barium nitrate solution corresponds to 0.001 ×0.1 = 0.0001 moles.

1 molecule of barium nitrate Ba(NO₃)₂ ​ will give 3 ions.

Hence, the number of ions present in 1 ml of 0.1 M Ba(NO₃)₂ solution = 0.0001 × 3 × 6.02 ×10²³ 

=1.8 × 10²⁰.

The angular momentumL = m_evr is on integer multiple of h/2π

mvr = nh/2π

for, n = 1

mvr = h/2π

True solution - Size < 1 nm

Colloidal solution - 1 nm < size < 1000 nm

Suspension - Size > 1000 nm

∵ ΔE = q + w

= 10 – 4 = 6 kJ

So, energy increases by 6 kJ

MA₂B₄ – 2 geometrical isomers

MA₄B₂ – 2 geometrical isomers

The complexes of general formula MA₆ and MA₅B having octahedral geometry do not show geometrical isomerism.

Fe²⁺ – 4 unpaired electrons

Ti²⁺ – 2 unpaired electrons

Cr²⁺ – 4 unpaired electrons

Hence the maximum number of unpaired electrons is present in Mn²⁺.

Magnetic moment ∝ number of unpaired electrons.

∴ 1 g of Al is obtained by charge of

3 × 96500/27 C.

∴ 5.12 × 10³ g of Al is obtained by charge of

= 3 × 96500/27 x 5.12 x 1000 = 5.49 × 10⁷C.

Thus, The integrated form for the first order reaction is C_t = C_oe^−Kt.

Here, C_o = represents initial concentration of the reactant, C_t represents the concentration of the reactant at time t and k is the rate constant of the reaction.

Other forms of the integrated rate law are ln C_t/C_o = -kt and k
=