Answer : Option BExplaination / Solution:
This is because the Nitro is electron withdrawing group,it pulls the electron density from the ring towards itself thereby decreasing the electron density in the ring and deactivating the ring towards attack by the electrophile.
Q3.The exhibition of various oxidation states by an element is also related to the outer orbital electronic configuration of its atom. Atom(s) having which of the following outermost electronic configurations will exhibit more than one oxidation state in its compounds.
Answer : Option CExplaination / Solution:
(3d24s2) is the configuration of transition element which shows variable oxidation state.
Q4.The order of screening effect of electrons of s, p, d and f orbitals of a given shell of an atom on its outer shell electrons is:
Answer : Option BExplaination / Solution:
Screening effect order follows s>p>d>f, this due to the nature of the orbitals involved. s-orbitals are least diffused and the two electrons in s-orbitals provide greatest shielding of the nuclear charge to the outermost electron. The diffused character increases in the order p>d>f. f-orbitals are so diffused that fourteen electrons inside them provide virtually no screening of the nuclear charge.
Q6.An atom of an element contains 29 electrons and 35 neutrons. The number of protons are:
Answer : Option BExplaination / Solution:
In a atom no. of protons = no. of electrons i.e. P = E while this is not true in case of ions.
So number of protons in given atom is 29.
Q7.Elements of which of the following group(s) of periodic table do not form hydrides.
Answer : Option CExplaination / Solution:
Elements of group 7, 8, 9 of d – block do not form hydrides at all. This inability of metal, of group 7, 8, 9 of periodic table to form hydrides is referred to as hydride gap of d – block.
Q10.The reducing power of a metal depends on various factors. Suggest the factor which makes Li, the strongest reducing agent in aqueous solution.
Answer : Option AExplaination / Solution:
High hydration energy of Li makes it’s a very good reducing agent. Its hydration energy compensates the high ionization energy.
Total Question/Mark :
Scored Mark :
Mark for Correct Answer : 1
Mark for Wrong Answer : -0.5
Mark for Left Answer : 0