Test: Periodic Properties- 1

The elememt belongs to group 2 and period 2 is Be( atomic number 4). Thus 2 electrons will be accommodated in first and other 2 in next shell(that is ,it is the outer or valence shell). Valence shell contains 2 electrons.

Nuclear mass does not affect the valence electrons because the nuclear mass is so small, it is considered as negligible. 

Nitrogen has higher ionization energy than Oxygen because it has a stable half-filled electronic configuration as shown below:
7 N - (1s)²(2s)²(2p_x)¹(2p_y)¹(2p_z) and

8 O - (1s)²(2s)²(2p_x)²(2p_y)¹(2p_z)¹

Isoelectronic species are the species belonging to different atoms or ions which have same number of electrons but different magnitudes of nuclear charges.

The size of an isoelectronic species increases with a decrease in the nuclear charge (Z). For example, the order of the increasing nuclear charge of F^–, Ne, and Na⁺ is as follows:

       F^– < Ne < Na⁺

Z     9     10     11

Therefore, the order of the increasing size of F^–, Ne and Na⁺ is as follows:

Na⁺ < Ne < F^–

When magnesium burns in air it forms magnesium nitride with nitrogen and with oxygen it forms magnesium oxide.

Be^- ions is most unlikely to exist as Be has higher tendency to loose electron and Be has full filled configuration.

Neutron/proton ratio is cause of radioactivity 

Electronic configuration of Ca is 1s²2s²2p⁶3s²3p⁶4s^2.
Filling of 3d orbital starts after the filling of 4s orbitals. If the aufbau principle had not been followed, the filling of 3d orbital would have been prior to filling of 4s orbital and the electronic configuration of Ca would have been  1s²2s²2p⁶3s²3p⁶3d² and it belongs to d block.

Electronic Configuration, which are given as follows:

• Electronic Configuration of element having Atomic number 33: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p³.
• Electronic Configuration of element having Atomic number 26: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁶
• Electronic Configuration of element having Atomic number 23:  1s² 2s² 2p⁶ 3s² 3p⁶ 4s²  3d³
• Electronic Configuration of element having Atomic number 15: 1s² 2s² 2p⁶ 3s² 3p³.

Observation of the above given 4 E.C. tells us that the element having Atomic number 33 will have the maximum number of unpaired electrons i.e. 3

1-it shows only +2 oxidation state- Hence False
2-is an f block element- Hence False
3-the bond formed is ionic because 1 is s block and ot is rb and 3 is a halide- Hence False
4-it is a d block element showing multiple oxidation states- Hence False

When period and group is same it can only be for 2s², and for 2s² principle quantum number (n)=2, azimuthal quantum number (l)= 0, magnetic quantum number (m)=0,spin quantum number (s)= +1/2 or -1/2.

The further away an electron is from the nucleus, the less force it feels from the electron, so the less energy is needed to “pop it off” the atom. The value of the energy level is exactly this amount of energy  so the smaller it is, the smaller the difference with neighboring levels will be.

Effective nuclear charge increases down the group and atomic radii increases due to addition of a new subshell

With increasing (+)ve oxidation state, acidity increases.

Na⁺: 1.02 A^o

Be²⁺: 0.39 A^o

Mg²⁺: 0.72 A^o

Li⁺: 0.76 A^o

The correct option is B.

F^- and Na⁺ both have 10 electrons .Atomic no of F is 9 and for Na it is 11.so for F^-, 9 proton is pulling 10 electrons and for Na⁺,11 proton is pulling 10 electrons . So attractive force is more for Na⁺ compared to F^-. So radius of F^- is more than Na⁺.

If the ions derived from different atoms are isoelectronic species, then they all have same number of electrons in their electronic shells and will have got same electronic configuration but their nuclear charge will differ because of their difference in number of protons in the nucleus. With increase in number of protonsin the nucleus the electrons are more attracted towards nucleus thereby causing the decrease in ionic radius. On this principle our problem will be solved

The given ions are

• 7N^3-→no. of proton=7 and no of electron=10
• ₈O^2-→no. of proton=8 and no of electron=10
• ₉F^-→no. of proton=9 and no of electron=10
• ₁₁Na⁺→no. of proton=11 and no of electron=10
• ₁₂Mg²⁺→no. of proton=12 and no of electron=10

Hence the increasing order of ionic radius is

₁₂Mg²⁺< ₁₁Na⁺< ₉F^-< ₈O^2-< ₇N^3-

For isoelectronic species lower the nuclear charge higher the radius

For Mg²⁺and Al³⁺ you can just check (z/e) ratio as these two are isoelectronic species, and you must know that, Li⁺ is slightly greater than Mg²⁺ ,this is a experimemntal data, the reason is out of your syllabi as it includes Schrodinger's eqⁿ,well K⁺ has the configuration of Ar[18] which makes it larger among them as noble gases are abnormally larger than their neighbor elements due to extra stability

If any atom loss electron then their radius will be decrease so , it means radius of atom K has more value then K⁺ ion
and if any atom gain electron then their size increases it means radius of F atom is smaller then F^- ion.