Test: Periodic Properties- 1

• The element belongs to group 2 and period 2 is Be( atomic number 4).
• Thus 2 electrons will be accommodated in the first and the other 2 in the next shell(that is, it is the outer or valence shell). The 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:
N - (1s)²(2s)²(2p_x)¹(2p_y)¹(2p_z)¹ 
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 the 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).
  Example: The order of the increasing nuclear charge of F^–, Ne, and Na⁺ is as follows: 
  F^– < Ne < Na⁺
  where Z = 9, 10, 11 respectively.
• Therefore, the order of the increasing size of F^–, Ne and Na⁺ is as follows:
  Na⁺ < Ne < F^–

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

Be^- ion is most unlikely to exist as Be has a higher tendency to lose an electron and Be has a full filled configuration.

Li^- =  1s², 2s² (In it all sub shells are saturated so, it is stable)

Be^- =  1s², 2s² 2p¹

B^-  =  1s², 2s² 2p²

C^-   =  1s², 2s² 2p³

• The electronegativities decrease from X to Y to Z.
• X, Y and Z could be phosphorus aluminium and sodium respectively.

The neutron/proton ratio is the cause of radioactivity. 

• Electronic configuration of Ca is:
  1s²2s²2p⁶3s²3p⁶4s²
• The 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 1s²2s²2p⁶3s²3p⁶3d and it belongs to d block.

The electronic configuration is 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.

Explanation: (I) [Kr]5s¹, shows only single oxidation state +1.
(II) [Rn]5f¹⁴ 6d¹ 7s², it is f-block element (Z=103).
(III) The compound formed between I and Ill is ionic.
(IV) [Ar]3d⁶ 4s², (Z=26) Fe shows variable oxidation state.
Therefore, all the above statements in the question are false.

When the period and group is the same, it can only be for Hydrogen(1s¹) and Beryllium(1s² 2s²). Then: 
Principal 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 neighbouring levels will be.

• Effective nuclear charge increases down the group and atomic radii increases due to the addition of a new subshell.
• Whereas on moving down the group, the force of attraction between the electrons and the nucleus decreases. Hence ionisation energy decreases.

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

In general, the ionic radius increases on moving from top to bottom in the group and decreases on moving from left to right in a period. 
The ionic radius of the following elements are:
Na⁺: 1.02 A^o
Be²⁺: 0.39 A^o
Mg²⁺: 0.72 A^o
Li⁺: 0.76 A^o
So, 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 protons are pulling 10 electrons and for Na⁺,11 proton are pulling 10 electrons. So attractive force is more for Na⁺ compared to F^-.
• So the radius of F^- is more than Na⁺.

• If the ions derived from different atoms are isoelectronic species, then they all have the same number of electrons in their electronic shells and will have got the same electronic configuration. 
• But their nuclear charge will differ because of their difference in the number of protons in the nucleus. 
• With an increase in the number of protons in the nucleus, the electrons are more attracted towards the 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 experimental 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 neighbour elements due to extra stability.

• If any atom loss electron then its radius will be decreased so, the radius of atom K has more value than the K⁺ ion.
• If any atom gain electron then its size increases, it means the radius of the F atom is smaller than the F^- ion.