16. BASED SOLELY ON ITS POSITION IN THE PERIODIC TABLE , PREDICT AND EXPLAIN HOW
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Question
16. BASED SOLELY ON ITS POSITION IN THE PERIODIC TABLE , PREDICT AND EXPLAIN HOW MANY 4D ELECTRONS ARE FOUND IN Z = 47?
17. PREDICT AND EXPLAIN WHICH ATOM WOULD HAVE THE HIGHER SECOND IONIZATION ENERGY: BARIUM OR CESIUM
18. USING THE SYMBOL FOR THE PREVIOUS NOBLE GAS TO INDICATE COMPLETELY FILLED PREVIOUS ROW ELECTRONS, WRITE THE ELECTRON CONFIGURATION FOR Z=50
19. USING VSEPR THEORY PREDICT AND EXPLAIN THE BOND ANGLE BETWEEN THE BROMINE ATOMS IN SBr2
20. for Z=28 which orbital is filled last? Why? Is this orbital unique?
21. explain why the metallic elements of a given horizontal row typically have much lower ionization energy than od the nonmetallic elements in the same row
Explanation / Answer
16) The electronic configuration of Z 47 is [Kr]5S^14D^10
We know that the element is more stable when it has completely filled or half filled configuration. So here 10 4D electrons present in Z 47 element (silver)
17) Cesium electronic configuration [Xe]6S^1
After first ionisation energy the electronic configuration becomes stable noble gas configuration. So it needs more ennergy to remove electron
Barium electronic cofiguration [Xe]6S^2
After first ionisation energy it reaches electronic configuration of Cesium. Thenn energy required to remove second electron is less compared to Cesium.
So finally the second ionisation energy is higher for Cesium than Barium
18) The electronic configuration of Z-50 is [Kr]4D^105S^25P^2 The element is Tin (Sb)
19)SBr2 the S undergoes SP^3 hybridisation with two lone pairs of electrons. According to VESPR theory it has bent structure like water molecule and the bond angle between two Br atoms is 104.45deg
20) Z=28 NickelElectronic configuration is 1S^22S^22P^63S^23P^64S^23D^8
According to Aufbau principle 4S orbital filled first than 3D.
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