Test : Chemistry Class 11 Mini Mock - 5 NEET MCQs & solutions - Free

3s is spherically symmetrical and its electron density is maximum at the nucleus. It decreases with r.

The balance combustion equation is C₂H₄ + 3O₂ -> 2CO₂ + 2H₂O, => x = 3, y = 2, => x + y = 5.

Since empirical formula is multiplied by n to get molecular formula.
CH₂O₂ will give only C₂H₄O₄ as its molecular formula. (CH₂O₂)_n where n = 1, 2, 3,... etc.

In anomalous Zeeman effect the splitting of spectral lines occurs due to interaction of the applied magnetic field with the magnetic moment generated by the orbital and 'spin' motions of the electron.

CH₃ – CH = CH₂ → CH₃ – CH⁺ – CH₂
The reason for this is only that carbocation is formed which has maximum stability. In this case, we have 6 α-H while for option a, b and d; we have 0, 2 and 2 α-H respectively. So only carbocation in option c forms.

Formula Mass is generally used for ionic compounds which do not contain discrete molecules, but ions as their constituent units.
The formula mass of a substance is defined as the sum of the atomic masses of constituent atoms in an ionic compound.

Chemical Formula for Ionic Compounds:

Only in CH₃CH₂OH, carbon has sp³ hybridisation. In other molecules, the carbon atom has multiple bonds.

Oxidation Numbers in Oxygen Difluoride (OF₂) and Dioxygen Difluoride (O₂F₂)

Oxidation numbers are assigned to elements in compounds according to several rules. Here, we'll use the rule that states the oxidation number of fluorine (F) in a compound is always -1, and the rule that states the sum of oxidation numbers of all atoms in a compound is 0.

Oxidation Number in OF₂:

• In Oxygen Difluoride (OF₂), there are two fluorine atoms, each with an oxidation number of -1. So, the total oxidation number contributed by fluorine is -2.
• To mak e the total oxidation number 0, the oxygen atom must have an oxidation number of +2.

Oxidation Number in O₂F₂:

• In Dioxygen Difluoride (O₂F₂), there are two fluorine atoms, each with an oxidation number of -1. So, the total oxidation number contributed by fluorine is -2.
• There are also two oxygen atoms. To make the total oxidation number 0, the combined oxidation number of the two oxygen atoms must be +2.
• Therefore, the oxidation number of each oxygen atom in O₂F₂ is +1 (because +2 divided by 2 equals +1).

So, in OF₂ and O₂F₂, the oxygen is assigned an oxidation number of +2 and +1, respectively. Therefore, the correct answer is B: +2 and +1.

Zinc gives H₂ gas with dil H₂SO₄/HCl but not with HNO₃ because in HNO₃, NO₃^– ion is reduced and give NH₄NO₃, N₂O, NO and NO₂ (based upon the concentration of HNO₃)

Zn is on the top position of hydrogen in electrochemical series. So Zn displaces H₂ from dilute H₂SO₄ and HCl with liberation of H₂.

There are 18 groups and 7 periods in periodic table. All members of 18^th group are noble gases and first group contains alkali metals which are the most electropositive.

Molecule has zero dipole moment, thus it does not have lone pair. Hence, non-polar.

Thus, (M — X) bonds indicate sp²-hybridisation.

Both XeF₂ & CO₂ have linear structures. Both are actually linear Tri-atomic molecules. 

Specific volume (volume of 1 gm) of cylindrical virus particle = 6.02 × 10^–2 cc/gm
Radius of virus (r) = 7 Å = 7 × 10^–8 cm
Length of virus = 10 × 10^–8 cm
Volume of virus

= 154 × 10^–23 cc

Wt. of one virus particle

∴ Mol. wt. of virus = Wt. of N_A particle

= 15400 g/mol = 15.4 kg/mole

Alkali metals typically form ionic bonds with non-metals like oxygen, resulting in ionic compounds such as oxides (e.g., Na₂O). They do not generally form covalent bonds with oxygen.

Therefore, statement A is false because alkali metals primarily form ionic bonds when reacting with oxygen, not covalent ones.

T₁ = T₂ because process is isothermal
Work done in adiabatic process is less than in by isothermal process as area under isothermal curve is more than adiabatic curve.
In adiabatic processes, expansion is done by using internal energy. Hence it decreases while in the isothermal process, temperature remains constant. So   there is no change in internal energy.

Higher the no. of electron releasing group lower will be stability of carbanion, and vice-versa. So the order of stability of carbanions is

The correct order for the relative energies of ethane conformations is as follows:

• Staggered (lowest energy)
• Skewed
• Eclipsed (highest energy)

This means that staggered conformations are the most stable, while eclipsed conformations are the least stable due to increased steric strain.

• Cisplatin is particularly effective against testicular cancer.
• Taxol (paclitaxel) is a mitotic inhibitor used in cancer chemotherapy.
• AZT is used for helping AIDS victims.

Additional Information: Its mode of action has been linked to its ability to crosslink with the purine bases on the DNA; interfering with DNA repair mechanisms, causing DNA damage, and subsequently inducing apoptosis in cancer cells. 

Attractive forces tend to bring two atoms close to each other whereas repulsive forces tend to move them away. In hydrogen, the magnitude of the new attractive forces is greater than that of new repulsive forces. As a result, two atoms come close to each other and potential energy decreases. The atoms approach each other until the equilibrium stage is reached where the net force of attraction balances the force of repulsion and system acquires minimum energy. 

In this configuration, there are four completely filled bonding molecular orbitals and one completely filled antibonding molecular orbital. So that N_b = 8 and
N_a = 2 .

∴ Bond order =

The correct answer is Option C.
From the image we can see that 3 isomers are possible for pentane.

Avogadro made the distinction between atoms and molecules, which today seems clear. However, Dalton rejected Avogadro's hypothesis because Dalton believed that atoms of the same kind could not combine. Since it was believed that atoms were held together by an electrical force, only unlike atoms would be attracted together, and like atoms should repel. Therefore it seemed impossible for a molecule of oxygen, O₂, to exist. Avogadro's work, even if it was read appears not to have been understood, and was pushed into the dark recesses of chemistry libraries and ignored. Avogadro continued to teach at the university of Turin, when it was not closed because of the political upheavals going on in Italy at the time, and died in 1854, an unknown figure. 

Work done in isothermal process, w = nRTln(V₂/V₁)
w₆₀₀/w₃₀₀ = 1×R×600×ln(10)/ 1×R×300ln(10) = 2
∆U = 0 for isothermal processes.

The element with atomic number(Z) 17 & mass number (A) 35 is chlorine = 

Combining the geometrical isomers with the orientations of the cyclobutyl substituent results in a total of 4 distinct stereoisomers for 1-chloro-2-(3-chlorocyclobutyl)ethene.

Answer:
B: 4

MnO₂ is used for the selective oxidation of allylic and benzylic carbon and in these compounds these carbons are not very easily oxidisable.