periodicity (chapter 7) and reactivity trend (chapter 8 )

Question 1

history of the periodic tabel

Answer 1

then: order of atomic mass and lined up with similar properties , gaps left , group trends were predicted now: 114 element, 7 horizontal periods

Question 2

how do chemicals have similar chemical properties?

Answer 2

atoms of each element have similar electron configuration

Question 3

trend across a period

Answer 3

period 2: 2s subshell filsl w 2 electrons, followed by 2p w 6 electrons period 3 : same pattern for 3s and 3p period 4: 3d subshell is involved, highest hell number is n=4. from n=4 shell, only 4s and 4p subshells occupied

Question 4

ionsiation energy graphs

Answer 4

each peak is a NOBLE GAS

Question 5

2p and 2s subshell

Answer 5

2p subshell has a higher energy than the 2s subshell therefore in boron the 2p electron= easier to remove than 2s electrons in beryllium . first ionisation energy of boron= less than beryllium

Question 6

nitrogen and oxygen

Answer 6

oxygen has repulsion in 2p orbitals- easier to remove =less first ionisation energy usual 2p susbehll, electrons repel as far apart as possible!

Question 7

Question 8

metal bonding

Answer 8

cations fixed in position

electrons mobile

2+ cations, twice as many delocalised electrons pressent

Question 9

diamond

Answer 9

109.5 degree bond angle

Question 10

giant covalent

Answer 10

high melting and boiling points- storng covalent obonds

solubility- insoluble, covant bonds too strong

electrical conductivit- not indiamond and silicon due to alll 4 outer shell electrons being used in cov bonding

Question 11

green box: graphene and graphite

Answer 11

bond angle 120

thinnest material ever made ( graphene )

layers bonded by weak london forces

Question 12

mp across periods

Answer 12

sharp decrease ebtween giant structures and simple molecules

. increases from group 1 to group 14, decreases sharply from group 14 to 15- due to change from giant simple molecular structure

giant molecular- strong forces to overcome

simple molecular = weak forces to overcome

Question 13

sequence of tests MEMORISE

Answer 13

1. carbonate test

.effervesncence from carbon dioxide gas, sulfate andh alide = no bubbles with dilute acid- if test produces no bubblles then carbonate not present

2. sulfate etst

add soltuion containing barium ions and look for white precip of baso4

.baco3 = white and insolluble in water- so if carry out sulfate test on carbonate–> get white precip which is hwy you carry carbonate test first so you know that NO CARBONATE IS PRESENT !

3. halide test

Add solution containing ag + ions , as agno3 .

ag2co3, silver carbonate and agso4= silver sulfate , both insoluble so carry put halide test last (so that the silver nitrate doesnt react w carbonates and sulfate and gives flase results )

Question 14

carrying out the chemical tests for mixtures of ions !

MEMORISE

Answer 14

1. carbonate test … see bubbles, continue adding nitric acid until stops

all carbonate ions will be removed and there will be none to react next tests

(if you itnend to test for sulfate or halide ions, use nitrc acid as sulfuric acid and hcl contain sulfate and halide ions)

2. sulfate test: add ba (no3)2 in excess to solutiion left from carbonate test . any sulfate ions come out as precipiatate of barium sulfate –> filter solution–> remove barium sulfate ( important not to use barium chlroide if you intend to test for halide ions next)

halide test

. to solution left from sulfate test, add ag no3

. any carbonate and sulfate ions initially present removed

. and nh3 to differentiate between precipitates and confirm halide

Question 15

test for ammonium ion

Answer 15

ammonium ion plus warm naoh –> ammonia gas released

no bubbles as ammonia very soluble in water