Chemistry

As 3f-orbital does not exist as value of l cannot be equal to n, hence no electron can be accommodated.

ΔH_rxn = ΔH_f (N? O, g) + 3ΔH_f (CO? , g) – (2ΔH_f NO? , g) – 3ΔH_f (CO, g)
= 81 + 3* (– 393) – 2 * 34 – 3 (–110)
= 81 – 1179 – 68 + 330 = – 836 kJ

The number of chiral carbons in open-chain aldohexose (such as glucose) is four, therefore, the number of stereoisomers = 2? = 16.

ATP has phosphate group, PO? ³?
So x + 4 (– 2) = – 3 or x = + 5

One mole of water is converted to vapour at its boiling point which is 100°C and at 1 atm. For this process ΔG = 0. As phase transformation of water is an equilibrium process and at equilibrium, free energy change is always zero.

Sometimes, when maximum covalency is obtained, the halides become inert to water, thus SF? (or similarly CCl? ) is stable. This is because SF? is coordinately saturated and sterically hindered. Thus, SF? is inert to water, because of kinetic rather than thermodynamic factor.
A. PCl? + 4H? O → H? PO? + 5HCl
B. SiCl? + 4H? O → Si (OH)? + 4HCl
C. BCl? + 3H? O → B (OH)? + 3HCl

2CO (g) + O? (g) → 2CO? (g)
Δn_g = 2 – (2 + 1) = -1
ΔH = ΔE + Δn_g RT or ΔH = ΔE – 1RT
i.e. ΔH < E

All monosaccharides which differ in configuration at C? and C? gives the same osazone. Since, glucose and fructose differ from each other only in configuration at C? and C? therefore, they give the same osazone. All other options given in the questions do not satisfy this condition and hence, do not from the same osazone.

The second ionization energy of K is maximum, because the second electron is removed from fully filled 3p? subshell. The second ionization energies of group 2 elements decrease down the group.
Hence, second I.E. of Ca > second I.E. of Ba.