IB DP Chemistry Topic 3.2 Periodic trends HL Paper 1

Question

Which element is in group 13?

 

Ionization energy / kJ mol–1

 

1st

2nd

3rd

4th

A.

789

1580

3230

4360

B.

578

1820

2750

11 600

C.

738

1450

7730

10 500

D.

496

4560

6910

9540

▶️Answer/Explanation

Ans: B

Successive ionisation energies increase because once you have removed the first electron you are left with a positive ion. Trying to remove a negative electron from a positive ion is going to be more difficult than removing it from an atom. Removing an electron from a 2+ or 3+ (etc) ion is going to be progressively more difficult.

The fourth ionisation energy is so large as compared to the third or lower ones. The electronic structure of Group 13 elements is All group 13 elements have ns2np1 valence electron configurations. The first three electrons to be removed are the three electrons in the np and ns orbitals. Once they’ve gone, the fourth electron is removed from the (n-1)p level – much closer to the nucleus. e.g. Aluminium is 1s22s22p63s23p1. The first three electrons to be removed are the three electrons in the np and ns orbitals. Once they’ve gone, the fourth electron is removed from the 2p level – much closer to the nucleus, and only screened by the 1s2 (and to some extent the 2s2) electrons. 

Hence, element B belongs to group 13.

Question

Which trend is correct, going down group 1?

A Melting point increases

B Reactivity decreases

C First ionisation energy increases

D Electronegativity decreases

▶️Answer/Explanation

Ans: D

As the metal atoms increase in size, any bonding electron pair becomes farther from the metal nucleus, and so is less strongly attracted towards it. This corresponds with a decrease in electronegativity down Group 1.

Question

Which species has the largest radius?

A.     Cl

B.     K

C.     Na+

D.     K+

▶️Answer/Explanation

B

As we move down the group number of shell increases and hence radii increases.

So correct order is  K+
The higher in energy an atom’s outermost electrons are, the further away from the nucleus they will be. This in turn implies a larger atomic radius.
Potassium has its outermost electron on the fourth energy level.  K:1s22s22p63s23p64s1
K+:1s22s22p63s23p6
Its outermost electrons are located on the third energy level. 
So, size of  K>K+.
Also, K>Cl
Now consider K+ and Cl
K+:1s22s22p63s23p6
 Cl:1s22s22p63s23p6

The two ions have the same electron configuration, which means that they are isoelectronic.

Both ions have their outermost electrons located on the same energy level, but they do not have the same ionic radius.

That happens because the number of protons each species has in its nucleus varies.

More specifically, potassium, which has an atomic number equal to 19, has 19 protons in its nucleus. Chlorine, on the other hand, only has 17.

This is important because in the case of the potassium atom, the outermost electrons will be attracted by the nucleus more.

In other words, K+ has bigger effective nuclear charge than Cl, which translates to a bigger net positive charge felt by the outermost electrons.

This will compress the energy levels a bit and make the ionic radius smaller for the potassium cation.

Therefore, the chloride anion will have the larger atomic radius.

So, K has largest radius among these. 

Question

Which property increases down Group 1?

A.     Atomic radius

B.     Electronegativity

C.     First ionization energy

D.     Melting point

▶️Answer/Explanation

A

Atomic radius increases down the group as n increases from top to bottom. 

Question

Which statement about the elements in group 7 is correct?

A.     \({\text{B}}{{\text{r}}_{\text{2}}}\) will oxidize \({\text{C}}{{\text{l}}^ – }\).

B.     \({{\text{F}}_{\text{2}}}\) has the least tendency to be reduced.

C.     \({\text{C}}{{\text{l}}_{\text{2}}}\) will oxidize \({{\text{I}}^ – }\).

D.     \({{\text{I}}_{\text{2}}}\) is a stronger oxidizing agent than \({{\text{F}}_{\text{2}}}\).

▶️Answer/Explanation

C

Oxidising power is in the order:I2<Br2<Cl2<F2 based on electrochemical series.

Hence, \({\text{C}}{{\text{l}}_{\text{2}}}\) will oxidize \({{\text{I}}^ – }\).

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